# Forces, Vectors, Work, Energy, Power

In all cases, resistive forces act to oppose motion. Therefore, unless a force is applied to balance the resistive force the object will slow down. In space, there are no resistive forces and objects will move at constant speed in a straight line unless another force acts.

**Newton’s First Law of Motion:**

• If the resultant force acting on a body is zero, it will remain at rest or continue to move at the same speed in the same direction.

• If the resultant force acting on a body is not zero, it will accelerate in the direction of the resultant force.

We define the Newton as the force needed to accelerate a 1 kg mass at 1 m/s^{2}. Therefore, we can write:

**Force (N) = mass (kg) × acceleration (m/s**^{2}**)**

**Newton’s Second Law **

**Newton’s Third Law **

Whenever an object experiences a force it always exerts an equal . . . . . and . . . . . opposite force on the object causing the force.

### Motion in Circles and Centripetal Forces

### Work

Work is the product of the force on a body and the distance it moves in the direction of the force

Work done = force x distance moved in the direction of the force

Work is done whenever energy is changed from one form into another.

SI unit is joule (J)

Work is a scalar quantity

### Gravitational Forces

### Energy

Energy is defined as the capacity to do work

SI unit is joule (J)

Energy is a scalar quantity

**Kinetic energy is the energy a body possesses due to its movement**Kinetic energy can be classified into

**Translational kinetic energy**: possessed by bodies in translational motion (eg moving train) =**1/2 mv**^{2}**Rotational kinetic energy**: possessed by bodies in rotational motion (eg rotating merry-go-round)

**Potential energy is the energy a body possesses due to its position or state****Potential energy**can be classified into:**Gravitational potential energy**: possessed by a body due to its position =**mgh****Elastic potential energy**: possessed by a body due to its strained state of being stretched or compressed

**Eg. A ball of mass 500g is moving at a velocity of 5m/s. What is the kinetic energy of the ball?**

kinetic energy = 1/2 mv^{2 }= 1/2 x 0.5 x 5 x 5 = 6.25 J

**Eg. Billy has a mass of 40kg. He runs up a flight of 20 steps, each of height 0.25m. Calculate his gain in gravitational potential energy**

gain in gravitational potential energy = mgh = 40 x 10 x (20 x 0.25) = 2000 J

### Principle of Conservation of Energy

States that energy can neither be created not destroyed but can be transformed from one form into another with no change in its total amount.

This means that the total energy input into a process is the same as the total energy output.

We can use a more sophisticated energy transfer diagram, called a Sankey diagram, to show this.

**Eg. A ball of mass 3kg is dropped from a height of 5m. **

**i. calculate the gravitational potential energy of the ball before it is dropped**

**ii. calculate the speed of the ball on hitting the ground**

**iii. if the ball bounces to a height of 3m, with what speed does it leave the ground?**

**iv. explain why the ball does not reach its original height when it bounces up again**

i. gravitational potential energy = mgh = 3 x 10 x 5 = 150J

ii. The kinetic energy of the ball on hitting the ground is equal to the ball's original gravitational potential energy so the kinetic energy of the ball on hitting the ground = 150J

If the ball hits the ground with speed v,

1/2 mv^{2 }= 150

v^{2} = (150 - 2)/3 = 100

v = 10ms^{-1}

iii. The kinetic energy of the ball on leaving the ground is equal to its gravitational potential energy on rising to its maximum height, that is 3m.

The gravitational potential energy of the ball 3m above the ground = 3 x 10 x 3 = 90 J

The kinetic energy of the ball leaving the ground = 90 J

If the ball leaves the ground with speed v,

1/2 mv^{2} = 90

v2 = (90 x 2)/3

v = 7.746ms^{-1}

iv. Because part of its kinetic energy is changed into other forms of energy like sound and heat when it hits the ground

**Eg. A pendulum bob of mass 0.5kg is moved sideways until it has risen by a vertical height of 0.2m. Calculate the speed of the bob at its**

**i. highest point**

**ii. lowest point**

i. at the highest point, the kinetic energy of the bob = 0

if the speed of the bob at its highest point is v,

1/2 mv^{2} = 0

1/2 x 0.5 x v = 0

v^{2} = 0

v = 0

ii. according to the principle of conservation of energy, the kinetic energy of at the lowest point is equal to the gravitational potential energy at the highest point.

If the speed of the bob at its lowest point is v,

1/2 mv^{2} = mgh

v^{2} = 2 x 10 x 0.2 = 4

v = 2 m/s

### Power and efficiency

Power is defined as the rate of doing work

**Power**= work done/time takenSI unit is watt (W)

**Efficiency**is the ratio of useful output energy to the total input energy or the ratio of useful power to the total input power.**Efficiency = (useful output energy / input energy) x 100%**

**Eg. A crane can lift a 200kg mass through a vertical height of 5m in 4s. Calculate**

**i. the power output of the motor driving the crane**

**ii. the efficiency of the motor if the power input is 5kW**

i. power output = work done/time taken = (200 x 10 x 5)/4 = 2500W

ii. efficiency of motor = (power output/power input) x 100% = (2500/5000) x 100% = 50%

### Friction

The net force that slows down moving objects

Acts in the opposite direction of motion of object

**1. Static friction**

related to objects which are not moving.

amount of force applied = amount of friction

**2. Moving friction**

applied force does not affect friction

it can be affected by surface or sudden change in mass

**Advantages of friction**

enables walking

brakes of vehicles

**Disadvantages**

reduce efficiency of machinery

energy wasted as heat

**Methods to reduce friction**

lubricants

ball bearings

-----> so that moving parts are made smoother

### Terminal velocity

The greater the velocity of an object, the higher the air resistance.

Terminal velocity occurs when the accelerating and resistive force on an object are balanced.

**Key ideas:**Drag/resistive forces on objects increase with increasing speed for objects moving through a fluid, e.g. air or water.

When accelerating and resistive forces are balanced, Newton’s First Law says that the object will continue to travel at constant velocity.

When an object reaches terminal velocity, the force of gravity and air resistance are balanced, the object falls at a constant speed and doesn’t accelerate.

Factors affected: Size, surface area, weight and nature of medium where object is flying.

If an object is falling through a vacuum, there would be no air resistance, thus acceleration is due to gravity alone.

### Inertia

Resistance of an object to change.

The greater the mass the more resistant it is.

An object at rest will remain at rest and an object at motion will remain at a constant speed with an absence of a resultant force.

### Energy, Work, Power

Examples of Energy Transformations Involving Electrical Devices and the Impact of Electricity on Society

Bonus: Renewable Energy Sources

Renewable energy resources are those that are not used up like fossil fuels.

They can be used on a large scale, mainly to generate electricity, or for individual buildings either to provide heating or to generate electricity.

All of these resources have advantages and disadvantages.

To use renewable resources effectively a combination of different resources must be used, both on a national and local scale.

## Worked examples

1. a stone of mass 5kg is dropped through a distance of 2.0m. Find the work done by the gravity on the stone.

Work done = force x distance = weight x distance = (5 x 10) x 2 = 100J

2. Calculate the work done against gravity in lifting a load of weight 50N through a vertical distance of 30cm

Work done = force x distance = weight x distance = 50 x (30/100) J = 15J

3. An object 25kg is moved 2m on a smooth horizontal surface. Find the work done by its weight

weight of object = 25 x 10 = 250N acting vertically downwards

work done by weight = 250 x 0 = 0J (because vertical distance moved = 0)

## MCQ Questions

**1. When a book of mass 2kg was pushed along the horizontal surface of the table, the friction force measured was 5N. When the book was pushed along the same table with a force of 9N, it moved with a constant**

a. acceleration of 2.0 m/s^{2}

b. acceleration of 25 m/s^{2}

c. speed of 2.0 m/s

d. speed of 2.5 m/s^{2}

**2. A balloon filled with gas has a total weight of 1800N. The balloon descends with a constant speed of 3 m/s. What is the resultant force acting on the balloon during descent?**

a. 0N

b. 600N

c. 1800N

d. 5400N

**3. A crane lifts a load of 8000N through a vertical distance of 20m in 4s. What is the average power during this operation?**

a. 100W

b. 1600W

c. 40000W

d. 640000W

4. A toy car A moving with a speed of 30 m/s has a kinetic energy of 900J. Another toy car B has twice the mass of toy car A. If toy car B moves with a speed of 15 m/s, what is the kinetic energy of toy car B?

a. 450J

b. 900J

c. 1800J

d. 3600J

**5. A 60W fluorescent lamp converts half the electrical energy supplied into light energy. How much light energy does it emit in 1 minute?**

a. 30W

b. 60W

c. 1800W

d. 3600W

**6. A electric motor is used to lift a 200N load through 3m in 5s. If the motor has an efficiency of 40%, what is the total electrical energy used by the motor in one second?**

a. 48W

b. 300W

c. 1200W

d. 3000W

7. A trolley of mass 1.5kg is placed on a smooth table. If a constant force of 6N acts on the trolley, the acceleration produced by the force will be

a. 0.25 ms-2

b. 4 ms-2

c. 4.5 ms-2

d. 7.5 ms-2

**8. An object of mass 2kg moves with uniform velocity when a constant force of 10N acts on it. When the force is increased to 20N, the acceleration will be**

a. 4 ms^{-2}

b. 5 ms^{-2}

c. 6 ms^{-2}

d. 10 ms^{-2}

**9. The weight of a rocket in outer space is zero because**

a. its mass becomes zero

b. there is no frictional force

c. there is no gravitational force

d. the rocket is stationary

**10. A ball of mass 0.2kg is thrown to a height of 15m. What is the change in its gravitational potential energy? (g=10N/kg)**

a. 0.3 J

b. 3.0 J

c. 7.5 J

d. 30 J

e. 75 J

**11. A boy pushes a toy cart along a level road and then lets it go. As the cart is slowing down, the biggest energy change is from**

a. chemical to heat

b. chemical to kinetic

c. heat to kinetic

d. kinetic to chemical

e. kinetic to heat

**12. A girl weighing 400N takes 4s to run up the stairs 3m high. What is her average speed?**

a. 0.75 m/s

b. 0.8 m/s

c. 1.25 m/s

d. 1.33 m/s

e. 12 m/s

13. How much potential energy does she gain? (from question 12)

a. 120 J

b. 200 J

c. 400 J

d. 1200 J

e. 2000 J

**14. A block of mass 2kg slides from rest through a distance of 20m down a frictionless slope 10m high. What is the kinetic energy of the block at the bottom of the slope? (g = 10ms-2)**

a. 20 J

b. 40 J

c. 200 J

d. 400 J

e. 800 J

**15. What are the main energy changes in a hydroelectric power station?**

a. electrical -> kinetic -> heat

b. heat -> electrical -> kinetic

c. kinetic -> light -> electrical

d. kinetic -> potential -> light

e. potential -> kinetic -> electrical

**16. An electric motor runs with a steady input of 250 V and 4 A while raising a load of 1000N. Assuming the motor and transmission to be 100% efficient, what time is taken to lift the load vertically through a distance of 10m?**

a. 1 s

b. 1.5 s

c. 4 s

d. 10 s

e 250 s

**17. No work is done by an object at rest because**

a. no force is acting on the object

b. no distance is moved

c. heat is not produced

d. friction is acting on the object

**18. A mass of 40g is raised vertically from the ground to a height of 50cm, the work done in lifting the mass is**

a. 0.02J

b. 20J

c. 0.2J

d. 2000J

**19. During free fall, work is done by**

a. frictional force

b. magnetic force

c. gravitational force

d. centripetal force

**20. Kinetic energy is transformed into gravitational potential energy when**

a. a raindrop falls from the sky

b. a rubber band is stretched

c. a stone is thrown upwards

d. a bullet is fired horizontally

**21. A hammer of a pile-driver is lifted to a height of 2m in 0.5s. If the mass of the hammer is 500kg, the power required for the lifting is**

a. 500W

b. 1000W

c. 2000W

d. 20000W

**22. A car travels at a constant speed of 10m/s. What is the power of the car if the total resistant forces acting on it is 400N?**

a. 1/40 W

b. 40W

c. 400W

d. 4000W

**23. A known force is applied to an object on a horizontal, frictionless surface. What property of the object must be known in order to calculate its acceleration?**

a. density

b. mass

c. surface area

d. volume

e. weight

**24. Which expression is used to calculate force?**

a. frequency x wavelength

b. mass x acceleration

c. power + time

d. pressure x area

e. work x distance

**25. Which of the following is a vector quantity?**

a. energy

b. mass

c. temperature

d. time

e. velocity

**26. When a force is applied to a body, several effects are possible. Which of the following effects could not occur?**

a. the body speeds up

b. the body rotates

c. the body changes direction

d. the pressure on the body increases

e. the mass of the body decreases

**27. A girl weighing 400N takes 4s to run up the stairs as shown in the diagram. What is her average speed?**

a. 0.75 m/s

b. 0.8 m/s

c. 1.25 m/s

d. 1.33 m/s

e. 12 m/s

**28. How much potential energy does she gain? (from qn 27)**

a. 120 J

b. 200 J

c. 400 J

d. 1200 J

d. 2000 J

**29. An electric motor can lift a weight of 2000N through a height of 10m in 20s. What is the power of the motor?**

a. 10 W

b. 1000 W

c. 2000 W

d. 4000 W

e. 400 000 W

**30. What are the main energy changes in a hydroelectric power station?**

a. electrical --> kinetic --> heat

b. heat --> electrical --> kinetuc

c. kinetic --> light --> electrical

d. kinetic --> potential --> light

e. potential --> kinetic --> electric

**31. A spiral spring has a natural length of 10.0cm. When a load of 5N is placed at one end while the other end is fixed on a hook, the length of the spring becomes 11.0cm. What is the new length of the spring if the load is 20N?**

a. 12.0cm

b. 14.0cm

c. 20.0cm

d. 44.0cm

**32. A body whose mass is 4kg, is placed on a frictionless surface. It is being pulled by a spring balance and the acceleration produced is 1m/s**^{2}**. What is the reading on the spring balance?**

a. 4N

b. 5N

c. 36N

d. 40N

**33. A body weighs 50N on earth where the acceleration due to gravity is 10m/s**^{2}**. When taken to the moon, where the acceleration due to gravity is 1.6m/s**^{2}**, the body would have a weight, in newtons, of**

a. zero

b. 8

c. 50

d. 80

**34. A parachutist, whose body and equipment have a total mass of 150kg, descends vertically through the air at a steady speed of 10m/s. Taking g = 10m/s**^{2}**, the resultant force acting on him in this descent is**

a. 1500N upwards

b. 150N upwards

c. 0N

d. 1500N downwards

**35. A man weights 600N. He runs up stairs of total height 4 metres in 3 seconds. How much power is exerted by the man?**

a. 450 W

b. 800 W

c. 2400 W

d. 7200 W

**36. When two forces are combined, the size of the resultant depends on the angle between the two forces. Which of the following cannot be the magnitude of the resultant when forces of magnitude 3N and 4N are combined?**

a. 1 N

b. 3 N

c. 7 N

d. 8 N

**37. A rock of mass 20kg is travelling in space at a speed of 6m/s. What is its kinetic energy?**

a. 60 J

b. 120 J

c. 360 J

d. 720 J

**38. A block of mass 6kg is pulled across a rough surface by a 54N force, against a friction force F. The acceleration of the block is 6m/s**^{2}**. What is the value of F?**

a. 9 N

b. 18 N

c. 36 N

d. 54 N

**39. A girl of weight 500 N runs up a flight of stairs in 10 s. The vertical height of the stairs is 5 m. What is the average power developed by the girl?**

a. 50 W

b. 100 W

c. 250 W

d. 1000 W

**40. When a block of wood of mass 2 kg is pushed along the horizontal flat surface of a bench, the friction force is 4N. When the block is pushed along the bench with a force of 10 N, it moves with a constant**

a. speed of 3 m/s

b. speed of 5 m/s

c. acceleration of 3 m/s^{2}

d. acceleration of 5 m/s^{2}

**41. A person exerts a horizontal force of 600 N on a box that also experiences a friction force of 200N. If it takes 4.0s to move the box 3.0m, what is the average useful power?**

a. 150 W

b. 300 W

c. 450 W

d. 600 W

**42. Which of the following best describes the useful energy change that takes place inside a mobile phone when sound is being produced?**

a. electrical energy --> sound energy

b. chemical energy --> electrical energy --> heat energy

c. chemical energy --> heat energy --> electrical energy + sound energy

d. chemical energy --> electrical energy --> sound energy

**43. A car is being driven up a slope at a constant speed. Which of the following describes the energy conversion of the system?**

a. chemical into kinetic

b. chemical into potential

c. kinetic into potential

d. potential into kinetic

**44. A car goes down a slope at constant speed. Which of the following describes the energy conversion?**

a. chemical into kinetic

b. chemical into potential

c. kinetic into potential

d. potential into heat

**45. What is the effect of the air resistance on a falling object?**

a. the speed of the object is reduced

b. the acceleration of the object is reduced

c. the distance travelled by the object is reduced

d. the direction of motion of the object is changed

**46. An object is moving due east at a constant speed of 5m/s before two equal and opposite forces of 10N each act on the object at the same time. The object will**

a. move with higher speed in the same direction

b. move with lower speed in the same direction

c. continue to travel at the same speed in the same direction

d. change the direction of motion and travel in the opposite direction

**47. A uniform rectangular board 8m x 2m is acted on by three forces on the edges. X is a pivot at the centre of the board.**

**What should be the value of F such that the board remains in equilibrium?**

a. 35N

b. 40N

c. 45N

d. 50N

**48. A man exerts a horizontal force of 500N on a box, which also experiences a frictional force of 100N. How much work is done against friction when the box moves a horizontal distance of 3m?**

a. 300J

b. 1200J

c. 1500J

d. 1800J

**49. A mass of 30kg is being pulled up a slope as shown. **

**What is the total work done in moving the box up the slope?**

a. 500J

b. 600J

c. 800J

d. 1400J

**50. A force is applied to an object on a surface with a frictional force of 2.0N. It produces an acceleration of 3ms-2. Which are the possible values of the applied force and the mass of the object?**

Force/N Mass/kg

a. 15 5

b. 6 2

c. 2 2

d. 17 5

**51. A trolley weighing 5.0N is pulled along a level bench by a horizontal force of 10N. The force of friction acting on the wheels of the trolley is 2.0N. What is the size of the resultant force causing the trolley to accelerate?**

a. 3.0N

b. 8.0N

c. 10N

d. 13N

**52. A bus has a total mass of 12 000kg. It moves along a horizontal stretch of road at a speed of 10m/s. It then accelerates, reaching a final speed of 30m/s after 16s. What is the size of the average resultant force acting on the bus when it is accelerating?**

a. 7500N

b. 9600N

c. 15000N

d. 22500N

**53. A boy pushes a toy cart along a road and then lets it go. As the cart is slowing down, the biggest energy change is from**

a. heat to kinetic

b. kinetic to heat

c. kinetic to potential

d. potential to heat

**54. At a height of 20m above the ground, an object of mass 4.0kg is released from rest. It is travelling at a speed of 20m/s when it hits the ground. The object does not rebound and the gravitational field strength is 10N/kg. How much energy is converted into heat and sound on impact?**

a. 40J

b. 80J

c. 800J

d. 1600J

**55. A car moves from rest with uniform acceleration along a horizontal road. After travelling a distance of 100m, it has kinetic energy equal to 200 000J. What resultant force is acting on the car?**

a. 100N

b. 1000N

c. 2000N

d. 20 000N

### MCQ Answers

1. a

2. a

3. c

4. a

5. c

6. b

7. b

8. b

9. c

10. d

11. e

12. c

13. d

14. c

15. e

16. d

17. b

18. b

19. c

20. c

21. d

22. d

23. b

24. b

25. e

26. e

27. c

28. d

29. b

30. e

31. b

32. a

33. b

34. c

35. b

36. d

37. c

38. b

39. c

40. c

41. b

42. d (energy is stored in the phone as chemical energy in the battery)

43. b ("constant speed" means no change in kinetic energy)

44. d (as above)

45. b (air resistance can never slow down a falling object but can only cause the falling object to have acceleration smaller than 10m/s2)

46. c (when a pair of equal and opp. force acts on the object, the forces cancel each other's efforts and therefore the resultant force acting on the object is zero.)

47. c

48. a

49. d

50. d

51. b

52. c

53. b

54. c

55. c

## Structured Question and Worked Solutions

**1. State briefly the energy changes in the production of electricity from**

**a. the burning of coal**

**b. wind-power**

**c. nuclear fission**

**Solution**

a. chemical energy -> heat and light energy -> latent energy of vaporization -> kinetic energy of steam -> kinetic energy of turbines -> kinetic energy of generator -> electrical energy

b. kinetic energy of air -> kinetic energy of windmill -> kinetic energy of generator -> electrical energy

c. nuclear energy of uranium -> heat energy -> latent heat of vaporization -> kinetic energy of steam -> kinetic energy of turbines -> kinetic energy of generator -> electrical energy

**2. A boy of mass 30kg runs up a flight of stairs to a floor which is at a height of 5.5m in 6.0s. Taking the weight of 1 kg = 10 N, calculate**

**a. work done by the boy against gravity**

**b. average power developed by the boy**

**Solution**

a. work done = weight x distance = 30 x 10 x 5.5 = 1650 J

b. average power = work done/time = 1650/6 = 275 W

**3a. Define force and state its SI unit**

**b. Two forces acting at a point have magnitudes 3 N and 4 N. By means of a diagram, show the lines of action of the forces when their resultant is**

**i. 7 N**

**ii. less than 7 N but more than 1 N,**

**iii. 1 N**

**c. Two forces of magnitudes 70 N and 50 N act at a point so that the angle between their lines of action is 40**^{o}**. By means of a scale diagram, determine the magnitude and direction of the resultant force acting at the point.**

**d. In a study on impact, a bullet of mass 50g penetrates a target and is brought to rest from an initial speed of 500ms**^{-1 }**in 0.2 s.**

**i. calculate the average deceleration of the bullet over the 0.2 s**

**ii. find the retarding force acting on the bullet during its impact with the target**

**Solution**

a. One newton is defined as the force that gives a 1 kg mass an acceleration of 1ms^{-2}.

SI unit: newton (N)

C.

di. average deceleration = (500 - 0)/0.2 = 2500 ms^{-1}

dii. retarding force = (50/1000) x 2500 N = 125 N

**4. A student Ben, starting at point P, walks due North for 1hr at a constant speed of 4.0km/h and then, at the same constant speed, walks 4.0km due East, finishing at a point Q. In the same total time but at a different constant speed, a second student Tom walks directly from P to Q. Determine**

**i. the total distance walked by student Ben**

**ii. the distance walked by student Tom**

**iii. the velocity of student Tom**

**Solution**

i. 8.0km

ii. 5.7km

iii. speed = 5.7/2 = 2.85km/h

velocity of student Tom is 2.85km/h at 45^{o }to the North

**5. A petrol-driven car accelerates from rest to its cruising speed along a straight level road.**

**i. state the principal energy changes in the car and its surroundings**

**ii. the car now climbs a slope with no change of speed. Explain whether the rate of petrol consumption will increase, stay the same, or decrease**

**Solution**

i. chemical potential energy -> kinetic energy of car -> internal energy gained by road and air

The chemical energy in the fuel is converted into kinetic energy of the car and internal energy gained by the air and road due to friction.

ii. rate of petrol consumption increases. When the car climbs a slope, it gains gravitational potential energy because of work done against gravity.

**6. The useful power output of a small dc motor is used to raise a load of 0.75kg through a vertical distance of 1.2m. The time taken is 18.0s. The voltage across the motor and the current through it are constant at 6.0V and 0.30A respectively. Assuming that the gravitational force on a mass of 1.0kg is 10N, calculate**

**i. the power input to the motor**

**ii. work done in raising the load**

**iii. useful power output developed by the motor**

**Solution**

i.power input to motor = IV = 0.30 x 6.0 = 1.80 W

ii. work done = force x distance = 0.75 x 10 x 1.2 = 9 J

iii. power output = work done/time = 9/18 = 0.50 W

**7. The figure below shows a simple pulley system. Calculate**

**i. the work done by the man in lifting the load**

**ii. the gravitational potential energy gained by the load**

**iii. the efficiency of the pulley system**

i. work done = 50 x 3 = 150J

ii. gravitational potential energy = 40 x 3 = 120J

iii. efficiency = (120/150) x 100% = 80%

**8. A motor boat travels due north at a steady speed of 3.0m/s through calm water in which there is no current. The boat then enters an area of water in which a steady current flows at 2.0m/s in a south-west direction as shown. Both the engine power and the course setting remain unchanged.**

**a. In the space below, draw a vector diagram showing the velocity of the boat and the velocity of the current. Use the diagram to find**

**i. the magnitude of the resultant velocity of the boat**

**ii. the angle between the due North and the direction of travel of the boat**

**b. Calculate the distance the boat now travels in 5.0 minutes**

**c. The mass of the boat is 3.0 x 10**^{3}** kg (3000 kg). Calculate the additional force which needs to be applied to give the boat an initial acceleration of 2.5 x 10**^{-2}** m/s**^{2}** (0.025 m/s**^{2}**)**

**Solution**

8ai. scale: 1cm = 0.5 m/s

magnitude of resultant velocity = 2.15 m/s

8aii. angle = 42^{o}

8b. distance = 2.15 x 5 x 60 = 645 m

8c. F = ma = 3000 x 2.5 x 10^{-2} = 75 N

**9. In a laboratory experiment, a small trolley was accelerated from rest by applying a small force to it. The distance travelled by the trolley was measured as 1.1 m in a time of 0.55 s. Calculate its average speed.**

**During this movement, the trolley was uniformly accelerating from rest. Calculate its speed after 0.55 s and its acceleration during this speed.**

**The mass of the trolley is 1.2 kg. What is the force producing this acceleration?**

**Solution**

average speed = distance / time = 11 / 0.55 = 2 m/s

speed after 0.55 s --> v = (2 x average speed) - u = 4 m/s

acceleration = (v - u) / t = (4 - 0) / 0.55 = 7.273 m/s^{2}

force = ma = 1.2 x 7.273 = 8.73 N

**10. In a crash test a car of mass 1500kg containing a dummy is driven into a rigid barrier at a speed of 15m/s. The recorded results showed that the interval between the first contact with the barrier and the car coming to rest was 0.12s.**

**a. calculate the average deceleration of the car over the 0.12s**

**b. find the retarding force, assumed to be constant, acting on the car**

**c. One of the man-shaped dummies used in the above test was strapped in place with a safety belt. The dummy was found to have moved forward 0.25m against the force exerted by the belt. Given that the kinetic energy of the dummy just before impact was 7870 J, calculate the average force which acted in the dummy as it was stopping.**

**d. Explain why it is an advantage for anyone riding in the car to be brought to rest steadily over this distance of 0.25m rather than abruptly.**

**Solution**

10a. given u = 15 m/s, t = 0.12 s, v = 0 m/s

v = u + at

0 = 15 + a(0.12)

a = -125 m/s^{2}

therefore the deceleration of the car is 125 m/s^{2}

10b. F = ma = 1500 x -125 = -1.875 x 10^{5} N

therefore the retarding force is 1.875 x 10^{5} N

10c. loss in kinetic energy = work done = force x distance

7870 = F x 0.25

--> F = 31480 N

10d. The force acting on the passenger will be much greater and hence drastic injuries could result if the passenger is brought to rest suddenly.

**11. A bricklayer lifts 12 bricks each weighing 20 N a vertical height of 1.2 m in 30 s. and place them at rest on a wall. Calculate**

**a. the work done**

**b. the average power needed**

**Solution**

11a. work done = total weight x height = 12 x 20 x 1.2 = 288 J

11b. average power needed = work done / time = 288 / 30 = 9.6 W

**12. A small, hard ball of mass 0.14 kg is thrown vertically upwards and reaches a height of 12 m above the point from which it is thrown.**

**Calculate the least energy which it must be given when thrown. (take the force of gravity on 1 kg to be 10 N)**

**On a windless day an inflated ball of much larger volume but having the same mass is propelled upwards with the same energy. It reaches a considerably smaller height. Explain briefly why this is so.**

**Solution**

energy required = work done = force x distance = 0.14 x 10 x 12 = 16.8 J

A larger volume means that the ball has a larger surface area so it will experience a larger resistance. Hence some energy is lost resulting in a smaller height.

**13. An athlete throws a javelin of mass 0.80 kg so that its centre of gravity is raised from a height of 2.0m above ground level at the moment of release, to a maximum height of 14.0 during its flight.**

**Calculate the energy to lift it against gravity to this height. (force of gravity on 1 kg is 10N)**

**Explain why the energy with which the javelin leaves the athlete's hand is considerably greater than the energy calculated above.**

**Solution**

vertical distance = 12m

Force = 0.8 x 10 = 8 N

work done = force x distance = 8 x 12 = 96 J

This is because the energy calculated is for the work done to lift the javelin vertically upwards. In the motion, the javelin also moves horizontally. Therefore extra energy is needed to do the work.

**14. A steady force of 6.0 N is applied horizontally to a body of mass 4.0 kg, which is initially at rest. In the 2.0 s during which the force is applied, the mass moves 3.0 m in the direction of the force. Assuming that there is no resistance to the motion, find**

**a. the work done by the force**

**b. the resulting kinetic energy of the body**

**c. the resulting velocity of the body**

**Solution**

14a. work done = force x distance = 6 x 3 = 18 J

14b. kinetic energy = work done = 18 J

14c, kinetic energy = 1/2 mv^{2}

18 = 1/2 x 4v^{2}

v = 3 m/s

**15. An empty lift is counterbalanced by a heavy piece of metal. Some people of combined mass 350 kg enter the lift and operate it. The lift rises 50 m in 60 s. Calculate**

**a. the work done in raising the people**

**b. the power required to do this**

**(take weight of 1kg to be 10N)**

**Solution**

15a. work done = force x distance = 3500 x 50 = 175 000 J

15b. power = work done / time = 175 000 / 60 = 2916.7 W

**16. A stunt man has one end of a thick elastic cord attached to him. The other end of the cord is firmly attached to a point on a high bridge. When the man jumps from the bridge he falls freely under gravity for 2.5s. Take the acceleration of free fall to be 10m/s2 and assume that the man is initially at rest.**

**a. Calculate**

**i. the vertical speed the man acquires during his free fall**

**ii. the vertical distance fallen**

**Suggest one reason why, in a real jump, the distance fallen in 2.5 s and the speed reached would be less than your calculated answers, even though the cord was slack throughout the 2.5 s.**

**b. After this time the cord begins to stretch and the man falls with continually reducing downward acceleration. Why is this?**

**c. Eventually his downward acceleration becomes zero. Explain why this happens.**

**If the mass of the man is 80 kg, suggest a value for the tension in the cord when his downward acceleration is zero.**

**Without making any further calculation, describe his motion after the point where his downward acceleration has become zero.**

**Solution**

16ai. vertical speed, v = u + at = 0 + 10(2.5) = 25 m/s

16aii. vertical distance fallen = (u + v) / 2 x 2.5 = (25/2) x 25 = 31.25 m

The existence of the air resistance brought about a smaller resultant force.

16b. As the stunt man falls further, the tension in the cord increases and together with air resistance reduce the resultant force.

16c. This is because the tension in the cord equals the man's weigh; no net forces is present at this time.

tension in the cord = weight of man = 80 x 10 = 800 N

The man begins to oscillate up and down about this point.

**17. A spring has a length of 5.0cm when it has no load hanging on it. When a load of weight 30N is hung from it, its length becomes 11.0cm. How long will it be if the weight of the load is changed to 20N?**

**Solution**

9.0cm

**18. A boy riding a bicycle has a total mass of 60kg and an acceleration of 0.6m/s**^{2}**.**** ****Calculate the accelerating force acting on the boy and the bicycle.**

**Solution**

36N

**19. A mass of 8kg is given an acceleration of**

**a. 5m/s**^{2}**.**

**b. 40cm/s**^{2}**.**

**What is the force acting in each case?**

**Solution**

19a. 40N

19b. 3.2N

**20. An object experiences 2 forces. A force of 3N pulls it horizontally to the right and one of 6N is applied at 60˚ to the horizontal. Draw a scale diagram to find the resultant and its direction**

**Solution**

7.9N at 41˚ to the horizontal

**21. Two girls Amy and Betty are pulling a rope with a ribbon attached to the rope as shown below. Amy is pulling the rope with a horizontal force of 550N while Betty is pulling the rope with a horizontal force of 700N. **

**a. Calculate the net force on the rope. In which direction will the ribbon move?**

**b. The two girls have the same mass of 35kg each and the rope and ribbon have a combined mass of 5kg. Calculate the acceleration of the ribbon.**

Solution

a. Net force - 700N - 550N = 150N

Direction: to the right, backwards from Betty.

b. F = ma

150N = 5 x a = 30m/s^{2}

**22. A 2.0kg mass has an initial speed of 2.0m/s. It is moved along by various forces as shown. For each scenario, sketch the velocity-time graph for the first 8.0s of its motion, showing appropriate workings where necessary.**

**Solution**

**23. Two masses, 20.0kg and 4.0kg, are connected to two ends of a rope that passes over a smooth pulley as shown in the diagram.**

**a. Mark and label on the diagram all the forces acting on the two masses.**

**b. The flat support is removed.**

**i. State the subsequent motion of the masses.**

**ii. Calculate the acceleration of the connected masses.**

**Solution**

*Note:*

*The four forces are:*

*2 x T**200N**40N**Normal reaction*

**24. The mass of a train is 2 x 10**^{6}** kg. The engine produces a forward force of 200kN. While accelerating, the average drag force is 80kN.**

**a. Calculate the average acceleration of the train.**

**b. The train enters a tunnel as shown below, travelling downhill (A), horizontally (B) and uphill again (C).**

**For safety reasons, the train must maintain a steady speed through the tunnel. Even though the speed remains steady, the driving force from the motors varies.**

**State whether the driving force from the engine motor is greater, less than, or equal to the drag force of 80kN along each of the following parts of the journey. Assume that the average drag force remains the same.**

**i. Downhill (A)**

**ii. Horizontal (B)**

**iii. Uphill (C)**

**Solution**

a. F = ma

(200 x 1000N) - (80 x 1000N) = (2 x 10^{6} kg) (a)

a = 0.060 ms^{-2}

bi. Driving force is smaller than 80kN

Note: The driving force, together with the weight of the train along the slope, will balance the average drag force.

bii. Driving force equals to 80kN.

biii. Driving force is greater than 80 kN.

*Note: The driving force must be greater than 80kN in order to balance the average drag and the weight of train along the slope.*

**25. Lily wants to hang a painting in a certain way. The diagram below shows what she has in mind.**

**a. Draw and label on the diagram all the forces acting on the painting.**

**b. The tension in the right string is 10N and the weight of the painting is 12N. By means of a scaled diagram, find the magnitude and direction ø of the tension of the left string.**

**Solution**

a.

b. Scale: 1cm : 1N

Tension = 5N

Ø = 35º

**26a. Define power and give its unit of measurement in SI units.**

**b. Coal can be burned to produce steam to rotate turbines which drive an electricity generator. When producing electricity, the rate of energy input to one generator is 1.2 x 10**^{13}** J/hr when the electrical power output is 1.2 x 10**^{-3}** W. Determine the efficiency (% of output over input) of the system working at this rate.**

**Answers**

12a. Rate of work done, J/s

b. 36%

**27. A trolley is released from rest on an inclined slope 1.5m high as shown. The motion is recorded by a strobe photograph as shown. The strobe frequency is 10Hz (ie it takes 10 pictures in 1s)**

**i. Determine the acceleration of the trolley down the slope from the figure on the left given that the mass of the trolley is 0.50kg and that it reaches the bottom after 0.55s.**

**ii. Determine the acceleration of the trolley from the figure on the right.**

**iii. Why are your answers in i and ii different?**

**Answers**

i. 10m/s^{2} (using principle of conservation of energy)

ii. 4m/s^{2}

**27. A box of mass 2kg achieves a constant speed of 4m/s when a 10N horizontal force is applied. **

**a. What is the friction force acting on the box?**

**b. Describe the motion of the box when the force increases from 10N to 12N.**

**Solution**

27a. Applied force - opposing force = ma

10 - F = 2 x 0

F = 10N

2b. Applied force - opposing force = ma

12 - 10 = 2 x a

a = 1m/s^{2}

The box has a constant acceleration of 1 m/s^{2}

**28. The figure below shows a man making a bungee jump.**

**The man starts his jump from a platform above a river. The elastic rope tied to his feet becomes tight when the man reaches point A. The lowest point he reaches is B. The mass of the man is 80kg. The gravitational field strength is 10 N/kg.**

**a. Describe the energy changes as the man falls from A to B.**

**b. The man falls 56m from the platform to point B. Calculate the energy change in his gravitational potential energy.**

**c. The maximum kinetic energy of the man is 23 000 J. Calculate his maximum speed.**

**Solution**

a. At A, the man has KE and PE. As he does down, his KE increases and PE decreases. Some KE of the man is changed into PE of the rope.

At B, acceleration is 0. KE is maximum as some PE is changed into KE.

b. mgh = 80 x 10 x 56 = 44 800J

c. 1/2mv^{2} = 23 000

v^{2} = 23 000/40 = 575

v = 23.98 m/s