Energy Conversion Q&A

8
Work, potential and kinetic energy

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good for student to have their self study.

Transcript of Energy Conversion Q&A

Page 1: Energy Conversion Q&A

Work, potential and kinetic energy

Page 2: Energy Conversion Q&A

1. What equation links work, force and distance? Work(J) =

force(N) X distance(m)

2. A force of 10N is applied to a box to move it 2m along the floor.

What is the work done on the box? 2x10 = 20J

3. If 50J of work is done on a box using 15N of force, how far

was the box moved? 50/15 = 3.3m

4. What equation is used to calculate increases in gravitational

potential energy? PE = mass x gravity x height OR PE =

weight x height

5. A crate of mass 100kg was lifted to a height of 75m. How

much work was done on the crate. Is this the same as the

gravitational potential energy that the crate now has?

Assuming gravity is 9.8m/s2. 100x9.8x75 = 73500 J. Yes

6. A book of weight 2N is lifted 1.5m onto a shelf. How much

potential energy has the book gained? 2x1.5 = 3J

7. What is the equation that links kinetic energy with mass and

velocity? KE = ½ mv2. (½ x mass x velocity2)

8. If an object of mass 15kg is travelling at 5m/s, how much

kinetic energy does it have? ½ x 15 x 25 = 187.5J

9. A car is travelling at 30m/s and has kinetic energy of 450kJ,

what is its mass? m=2xKE/v2=2x450000/900=1000kg

10. A tennis ball has a mass of 0.06kg and has kinetic energy of

2.94kJ when served, what is its speed? v=(2KE/m)

=(5.88/0.6)=3.13m/s

Page 3: Energy Conversion Q&A

A forklift truck was used to stack boxes on a trailer.

It lifted a box weighing 1900N through 4.5m.

(a) Calculate the work done on the box. Show your working.

…work=force x distance = 1900X4.5 =

8550 ………………………………………Work done

= …8550…..J

(3 marks)

(b) The efficiency of the forklift truck is about 80%.

Explain what happens to the wasted energy?

……the motor, due to friction, has heated up and it is also noisy

this energy has been lost to the surroundings…

(1 mark)

Page 4: Energy Conversion Q&A

The diagram shows how far a dummy in a car crash test moves

before it is stopped by the seat belt. The graph shows the force

exerted by the seat belt on the dummy during the crash.

(i) Write down the equation that links distance moved, force

applied and work done.

.........work = force x distance......................

(1 mark)

(ii) Calculate the work done to stop the dummy.

Show clearly how you work out your answer.

.....................work = 7000 x .03

= .....................................................................................................

....

Work done = .........2100............... joules

(2 marks)

(iii) How much energy is transferred from the dummy during the

crash?

..............2100 J........................................................................

(1 mark)

Page 5: Energy Conversion Q&A

The diagram shows a tower used for bungee jumping.

(a) A lift takes a person of mass 55 kg, to a platform 64 metres

above the ground.

(i) Write down the equation that links change in gravitational

potential energy, change in vertical height and weight.

..................PE = weight x change in vertical height ....... (1 mark)

(ii) Calculate the increase in the gravitational potential energy of

the person when going from the ground to the platform.

Show clearly how you work out your final answer.

(Gravitational field strength = 10 N/kg)

..................PE = 55 x 10 x 64

= ........................................................... Increase in

gravitational potential energy ....35200.. joules

(2 marks)

Page 6: Energy Conversion Q&A

(b) The falling bungee jumper reaches a maximum speed when

15 840 joules of gravitational potential energy have been

transferred into kinetic energy.

(i) Write down the equation that links kinetic energy, mass and

speed.

...............KE = ½ mv2......................... (1 mark)

(ii) Calculate the maximum speed of the falling bungee jumper.

Show clearly how you work out your final answer.

.....................v = (2KE/m) = (2x15840/55) =

(576) ....................................................................................

Speed = ........24.................. m/s

(3 marks)

(c) As the rope stretches the bungee jumper slows down.

Describe fully the energy transfer that happens as the bungee

jumper continues to fall towards the ground.

To gain full marks in this question you should write your ideas in

good English. Put them into a sensible order and use the correct

scientific words.

.....From maximum KE the bungee starts to stretch and the KE in

transferred into elastic potential energy... (3 marks)

Page 7: Energy Conversion Q&A

A pile driver is used to push steel supports into the ground. A pile

driver does this because it has a lot of kinetic energy as it hits the

steel support.

(a) Give two ways in which the kinetic energy of the pile driver as

it hits the steel support could be increased.

1 increase the mass of steel

2 increase the height from which it was dropped

(2 marks)

(b) The mass of the pile driver is 500 kg.

The kinetic energy of the pile driver as it collides with the steel

support is found to be 36 000 J.

Calculate the speed of the pile driver just before it hits the steel

support.

Show clearly how you work out your answer and give the unit.

KE = ½ mv2

m=500

KE = 36000 sqrt(2x36000/500) = sqrt(144)

........................................................................................................

........................................................................................................

Speed of the pile driver = ....12m/s................

(4 marks)

Page 8: Energy Conversion Q&A

A car which is moving has kinetic energy.

The faster a car goes, the more kinetic energy it has. The kinetic

energy of this car was 472 500 J when travelling at 30 m/s.

Calculate the total mass of the car.

Show clearly how you work out your answer and give the unit.

.........................................................................................................

KE = ½ mv2

.........................................................................................................

m = 2KE/v2

M = 2 x 472500/302 ............................................................

.........................................................................................................

.........................................................................................................

Mass of the car = ..1050 kg...............

(5 marks) (4 marks for getting to the correct answer, 1 mark for

the unit)