SITEMAP   School Physics: Energy transfers Section 0.4 Work done & energy stores

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Energy: Part 0.4 Mechanical work calculations, power, energy stores and energy transfers

Doc Brown's Physics exam study revision notes

0.4A Introduction to work done calculations, power and wasted energy

If an object begins to move or changes how it moves (change in speed or direction) then a force must be involved.

If a force acts on an object, then work is being done on the object increasing its energy store - this must involve an energy transfer.

If an object exerts a force or transfers energy in some way, work is done by this object and energy is transferred from its energy store.

If there is no friction and a force is applied to an object, the work done on the object will equal the energy transferred to the object's kinetic energy store.

However, if an object is already moving and experiences a resistive force of friction slowing it down, the energy transferred from its kinetic energy store is equal to the work done against the object's motion.

The energy can be transferred usefully i.e. doing useful work, but some energy might (often) be wasted - often dissipated to the surroundings as heat due to friction.

The energy transferred, usefully or wastefully, is often referred to, and equal to, work done.

When dealing with the action of a resultant force acting through a linear distance in the direction of the force, you can calculate the work done with a very simple formula. When ever you or a machine moves something work is done

Work done (J) = energy transferred = force (N) x distance (m)

W (J) = F (N) x d (m)

The distance equals the length of the line of action through which the force acts.

(work done  might be denoted by E or W for the energy of work done, but remember the unit for power is the watt, W, so take care!)

If you apply a force of one newton through a linear distance of one metre, you do one joule of work.

one joule of work done = a force of one newton x one metre distance, 1 J = 1 Nm

Imagine a force of 1 newton acting through a distance of 1 metre, 1 joule of work is done = 1 joule of energy transferred.

(note that newton metres (Nm) = joules (J)  = work done = energy transferred)

The distance must be along the line of action of the force.

You need an energy source to keep applying any force for any length of time.

Whenever work is done e.g. mechanically or electrically, energy is transferred from one energy store to another.

See Power is the rate at which energy is transferred or 'used up' - work done!

In other words power is the rate of doing work.

The unit of power is the watt, W.

power (W) = energy transferred (J) / time taken (s)

power (W) = work done (J) / time taken (s)

P (W) = E (J) / t (s)

{and don't forget: work done (J) = force (N) x distance (m), which you need to solve some power problems}

A power rating of one watt means one joule of energy is transferred per second,

or, 1 joule of work is done in 1 second.

The joule is the unit of energy and the watt is the unit of power (W).

one watt = one joule of energy is transferred per second, 1 W = 1 J/s

A 1 kW electric heater converts 1000 J of electrical energy  ==> heat energy every second

(1 kW = 1000 W  =  1000 J/s)

The power of a machine or any other device is the rate at which it transfers energy.

A powerful machine doesn't necessary mean a bigger force is generated, it just means a lot of energy is transferred in a short time (or a lot of work done in a short time).

The more work done/energy transferred in a given time, the greater the power.

The shorter the time taken to transfer energy/do work, the greater the power generated.

Conversely, by using gears and a low powered electric motor, you can generate a relatively big force. So, don't confuse power and force - use the terms appropriately.

See

However, no mechanical device or electrical appliance is 'perfect' and energy is lost e.g. by thermal energy by friction or sound from unwanted vibrations, heat loss from circuit wires etc., but it is still part of the total work done - it just isn't all useful!

Therefore you should know and understand that when energy is transferred only part of it may be usefully transferred as useful energy or useful work, the rest is ‘wasted’ to the surroundings

You should know and understand that wasted energy is eventually transferred to the surroundings, which will become warmer AND ...

... the wasted energy becomes increasingly spread out and so becomes less useful.

For more on wasted energy see:

0.4B More examples of energy stores, energy transfers and mechanical work calculations

Doing work in a mechanical sense is just another set of examples of energy transfer e.g. from one energy store to another.

This usually involves doing useful work, but the final energy store destination might have no further use.

For example when you burn fuel in a car engine the energy changes are as follows.

fuel + oxygen (chemical energy store) ==> hot gases (thermal energy store) ==> car moves (kinetic energy store)

In the process some of the original chemical energy store of the system is lost through friction and sound.

BUT, all the energy from the chemical energy store ends up as thermal energy to warm up the surroundings. Therefore the thermal energy store of the environment is increased but its now of no use whatsoever!

The waste of energy increases (but necessarily!) when you apply the brakes. Mechanical work is being done in the process - work is being done against the force of friction. The friction effect of the brake pad on the brake disc converts kinetic energy into thermal energy. The kinetic energy store of the car decreases (fortunately!) and the thermal energy store of the brake pad + disc increases. Eventually as the braking system cools down the heat energy is transferred to increase the thermal energy store of the surroundings - wasted or degraded energy!

If a vehicles crashes into a stationary object, the contact force causes energy to be mechanically transferred from the vehicle's kinetic energy store to elastic potential energy store of the crushed vehicle parts, the thermal energy stores of the vehicle, object crashed into and the surrounding air (including some sound energy too - which also ends up as thermal energy!).

When you lift an object up you are doing work against the effects of gravity.

The chemical energy stored in your muscles is decreased as some of it used to lift the object-weight which on gaining height increases its gravitational potential energy store.

In fact the work you do (J) equals the weight of the object (N) x vertical distance lifted (m)

When you kick a ball up in the air, three energy stores should be fairly obvious to you.

chemical energy store of muscles ==> kinetic energy store of ball ==> gravitational potential energy store of ball

AND, when the ball falls (ignoring a bit of lost sound energy)

gravitational potential energy store of ball ==>`kinetic energy store of ball ==> thermal energy store of the ground

For more on wasted energy see:

0.4C Examples of simple mechanical work done calculation questions

and Examples of simple power calculation questions How to solve 'work done' problems.

The formula for work done is quite simple:

work done in joules = force in newtons x distance in metres along the line through which the force acts

work done (J) = force (N) x distance through which force acts (m)

E  = F x d

In real situations of energy transfers involving mechanical work, I'm afraid friction effects accompany the useful work done.

Moving parts rubbing against each other causes friction and rise in temperature - kinetic energy store ==> thermal energy store of machine and surroundings.

Note that the force can be the weight of an object acting in a vertical direction ... see the page on ... How to solve power problems - just a few simple 'mechanical' examples here..

Power is the rate at which energy is transferred, that is the rate at which work is done.

power in watts = (work done = energy transferred) ÷ time taken

P (W) = E (J) / t (s)

Power P in Watts, Energy E in Joules, time t in seconds

The power rating of 1 watt is equal to a work rate of 1 joule of energy transferred per second

See also FORCES 3. Calculating resultant forces using vector diagrams and work done

Examples of problem solving using the 'work done' and 'work rate = power' formulae

In every case watch the units e.g. W/kW, J/kJ/MJ, min/secs etc. so take care!

Q1 If you drag a heavy box with a force of 200 N across a floor for 3 m,

(a) what work is done?

(b) If you do the job in 5 seconds, what was your power rating in doing this task?

Q2 (a) If a machine part does 500 J of work moving linearly 2.5 m, what force was applied by the machine?

(b) If an electric motor transfers 12.0 kJ of useful energy in 3.0 minutes.

Calculate the power output of the motor.

(c) An appliance has a power rating of 1.2 kW.

How many kJ of energy is transferred in 8.0 minutes?

Q3 A machine applies a force of 200 N through a distance of 2.5 m in 2.0 seconds.

What is the power of the machine?

Q4 A 500 kg express skyscraper lift moves non-stop up a total height of 100 m.

(a) If the force of gravity = 9.8 N/kg, calculate the weight of the lift.

(b) Calculate the work done by the lift motor.

(c) If the lift ascent time is 20 seconds, what is the power of the lift motor?

(d) What assumptions has been made for calculations (b) and (c)?

Q5 Part of a machine requires a continuous force of 500 N from a motor to move it in a linear direction.

(a) How much work is done in moving it a distance of 50 m?

(b)  If the power of the machine is 5.0 kW, how long will it take to move the machine part the 50 m distance?

Q6 A machine has to move a conveyor belt at a rate of 30 m/min.

The system is designed to use 6 kJ/min of electrical energy to move the conveyor belt along.

What is the minimum force the motor must produce to move the conveyor belt along?

Difficult question, involving 3 steps.

Q7 A toy model car has a clockwork motor, whose spring can store 8.75 J of elastic potential energy.

On release the clockwork motor can deliver a continuous force of 2.5 N.

Q8 A sliding object moving across a very rough surface has 5.0 J in its kinetic energy store.

If the object experiences a constant resistive force of friction of 25 N, calculate in cm how far the object travels before coming to a halt.

Q9 A small electric motor uses 120 J of electrical energy in 3.0 minutes.

Calculate the power of the electric motor.

Q10 A 45.5 kg mass is hoisted up vertically 15.5 m.

If the gravitational field force is 9.8 N/kg, calculate the work done to the nearest joule.

Q11 A small electric motor operating a water pump for a garden ornament does 6.5 kJ of work in 3.0 minutes.

Calculate the power of the pump.

Q12 An electrical appliance transfers 8000 J of energy in 1 minute and 40 seconds.

Calculate the power of the appliance.

Q13 A person weighing 600 N strides up 20 steps, each one 20 cm high, in 10 seconds.

(a) Calculate the work done.

(b) calculate the average power the person generates in the process of ascending the 20 steps.

(c) If the person stands still on the 20th step, what is the person's power?

Q14  (you might not have dealt with all aspects of Q14 yet)

Imagine a car of 1000 kg travelling at 20 m/s doing an emergency stop in a distance of 25 m - the braking distance.

Calculate the average braking force produced by the driver when pressing on the brake pedal.

To solve this question you to use several formulae.

(a) Calculate the kinetic energy of the car.

(b) What work must be done to bring the car to a halt? Explain your answer.

(c) Calculate the average braking force required.

Q15 A large car is travelling at 120 km/hour and the engine provides a constant driving force of 1500 N.

(a) How far does the car travel in 1 second?

(b) From your answer to (a) calculate the power of the engine.

Difficult question, you have to do a little 'think' connecting two equations.

Keywords, phrases and learning objectives on energy

Be able to do mechanical work done calculations involving power, force, energy stores and energy transfers.

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ANSWERS to the problem solving questions on work done and power

Q1 If you drag a heavy box with a force of 200 N across a floor for 3 m,

(a) what work is done?

work done = 200 x 3 = 600 J

(b) If you do the job in 5 seconds, what was your power rating in doing this task?

power (W) = work done (J) / time taken (s)

your power = 600 / 5 = 120 W  (just over the power of 100 W light bulb!)

Q2 (a) If a machine part does 500 J of work moving linearly 2.5 m, what force was applied by the machine?

work done = force x distance, rearranging,  force (N) = work done (J) ÷ distance (m)

force = 500 ÷ 2.6 = 200 N

(b) If an electric motor transfers 12.0 kJ of useful energy in 3.0 minutes.

Calculate the power output of the motor.

work done = energy transferred = 12 x 1000 = 12,000 J, time taken = 3 x 60 = 180 s

power = work done / time = 12,000 / 180 = 66.7 W (3 sf)

(c) An appliance has a power rating of 1.2 kW.

How many kJ of energy is transferred in 8.0 minutes?

1.2 kW = 1200 watts, time = 8 x 60 = 480 seconds

P = E / t, rearranging gives E = P x t

energy transferred = 1200 x 480 = 576 000 J = 576 kJ

Q3 A machine applies a force of 200 N through a distance of 2.5 m in 2.0 seconds.

What is the power of the machine?

work done = force x distance = 200 x 2.5 = 500 ?

power = work done / time taken = 500 / 2.0 = 250 J/s = 250 W

Q4 A 500 kg express skyscraper lift moves non-stop up a total height of 100 m.

(a) If the force of gravity = 9.8 N/kg, calculate the weight of the lift.

weight = mass x g

= 500 x 9.8 = 4900 N

(b) Calculate the work done by the lift motor.

work = force x distance

the force exerted by the lift motor must be at least equal to the weight of the lift due to gravity

therefore work done = 4900 x 100 = 490000 J 490 kJ 0.490 MJ

(c) If the lift ascent time is 20 seconds, what is the power of the lift motor?

power = work done / time taken

power = 490000 / 20 = 24500 W 24.5 kW

(d) What assumptions has been made for calculations (b) and (c)?

Both (b) and (c) calculations ignore the extra work done in overcoming any forces of friction.

Q5 Part of a machine requires a continuous force of 500 N from a motor to move it in a linear direction.

(a) How much work is done in moving it a distance of 50 m?

work done = force x distance = 500 x 50 = 25000 J (25 kJ)

(b)  If the power of the machine is 5.0 kW, how long will it take to move the machine part the 50 m distance?

power = work done / time taken

time taken = work done / power

time = 25000 / 5000 = 5.0 s (stand clear!)

Q6 A machine has to move a conveyor belt at a rate of 30 m/min.

The system is designed to use 6 kJ/min of electrical energy to move the conveyor belt along.

What is the minimum force the motor must produce to move the conveyor belt along?

(i) power = rate of energy transfer = energy transferred (J) / time taken (s)

power = 6000 / 60 = 100 W (100 J/s)

(ii) 30 m/min ≡ 30 / 60 = 0.5 m/s

so in one second, the energy transferred is 100 J and the distance moved is 0.5 m.

work = force x distance

force = work / distance = 100 / 0.5 = 200 N

(iii) If you are smart, you can solve the problem directly, because both bits of data involved 'per minute'.

So you can just say force = work / distance = 6000 / 30 = 200 N !!!

BUT, always work logically in your own comfort zone, the data might not always be so kind!

Q7 A toy model car has a clockwork motor, whose spring can store 8.75 J of elastic potential energy.

On release the clockwork motor can deliver a continuous force of 2.5 N.

How far will the car travel in one go?

energy store = total work done = force x distance

distance = energy store / force = 8.75 / 2.5 = 3.5 m

Q8 A sliding object moving across a very rough surface has 5.0 J in its kinetic energy store.

If the object experiences a constant resistive force of friction of 25 N, calculate in cm how far the object travels before coming to a halt.

energy transferred = kinetic energy = resistive force x distance to come to a halt

E = KE = F x d,  5.0 = 25 x d, d = 5.0/25 = 0.20 m = 20 cm

Q9 A small electric motor uses 120 J of electrical energy in 3.0 minutes.

Calculate the power of the electric motor.

Energy transferred = 120 J, time = 3 x 60 = 180 s

P = E / t  =  120/180  = 0.67 W (2 sf)

Q10 A 45.5 kg mass is hoisted up vertically 15.5 m.

If the gravitational field force is 9.8 N/kg, calculate the work done to the nearest joule.

weight = force = mass x gravity = 45.5 x 9.8 = 445.9 N

This force acts through the vertical height the object is lifted.

work done = force x vertical distance = 445.9 x 15.5 = 6911.45

work done = 6911 J

Q11 A small electric motor operating a water pump for a garden ornament does 6.5 kJ of work in 3.0 minutes.

Calculate the power of the pump.

Work done = 6.5 x 1000 = 6500 J

Time taken = 3 x 60 = 180 s

Power = work done / time taken = 6500 / 180 = 36.11 = 36 W (2 sf).

Q12 An electrical appliance transfers 8000 J of energy in 1 minute and 40 seconds.

Calculate the power of the appliance.

power = energy transferred / time = 8000 / 100 = 80 W

Q13 A person weighing 600 N strides up 20 steps, each one 20 cm high, in 10 seconds.

(a) Calculate the work done.

Each step is 20 cm = 0.20 m, total height = 20 x 0.20 = 4.0 m

Work done (J) = force (N) x distance (m)

Work done = 600 x 4.0 = 2400 J

(b) calculate the average power the person generates in the process of ascending the 20 steps.

power (W) = work done (J) / time taken (s)

power = 2400 / 10 = 240 W

(c) If the person stands still on the 20th step, what is the person's power?

Since no force is acting through a distance, in a mechanical sense, the power is zero, but ....

YES. Your body must be still releasing energy through respiration to maintain the tension in your muscles to keep you upright, otherwise you would flop down in a heap!

Q14  (you might not have dealt with all aspects of Q14 yet)

Imagine a car of 1000 kg travelling at 20 m/s doing an emergency stop in a distance of 25 m - the braking distance.

Calculate the average braking force produced by the driver when pressing on the brake pedal.

To solve this question you to use several formulae.

(a) Calculate the kinetic energy of the car.

KE = 0.5 mv2 = 0.5 x 1000 x 202 = 200 000 J

(b) What work must be done to bring the car to a halt? Explain your answer.

If the kinetic energy of the car is 200 000 J, then 200 000 J of work must be done to bring the KE of the car to zero i.e zero velocity.

(c) Calculate the average braking force required.

Work (J) = force (N) x distance (m)

work = 200 000 J and the braking distance was 25 m

force = work / distance = 200 000 / 25 = 8000 N average braking force.

Q15 A large car is travelling at 120 km/hour and the engine provides a constant driving force of 1500 N.

(a) How far does the car travel in 1 second?

1 km = 1000 m, 1 hour = 60 x 60 = 3600 s

speed (m/s) = distance (m) / time (s)

speed of vehicle = 120 x 1000 / 3600 = 33.33 m/s

so, in one second the vehicle moves 33.3 m (3 sf)

(b) From your answer to (b) calculate the power of the engine.

You have to do a little 'think' connecting two equations.

work = force x distance and power = energy transferred / time

i.e. energy transferred = work done = force x distance

power (W) = work done / time = force (N) x distance (m) / time (s)

power = 1500 x 33.33 / 1 = 50 000 W or 50 kW (3 sf)

In other words the distance / time in the equation is quite simply the speed of 33.3 m/s, get it OK?

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