MathsFurther Mechanics 1 › Work, energy and power

Work, energy and power

An accounting method rather than a new law. Add up what went in, subtract what was wasted, and whatever is left has to be somewhere.

Year FMEDEXCEL 9FM0 FM1

Builds on Forces and Newton's laws and Friction and inclined planes.

IN THIS TOPIC

  • Calculate work done by a force at an angle, and by gravity and friction.
  • Apply the work-energy principle to motion on a slope.
  • Use P = Fv to link engine power, driving force and acceleration.

WHAT YOU PROBABLY THINK

The work done by a force of 20 N moving an object 5 m is always 100 J.

Work, and where it goes

Work is the component of force along the displacement, times the displacement: Fd cos θ, where θ is the angle between them. A force at right angles to the motion does no work at all, which is why the normal reaction never appears in an energy equation. The opening claim forgets the angle, and would also be wrong for a force partly opposing the motion.

KE = 12mv2,      GPE = mgh

The work-energy principle says the total work done by all the forces equals the change in kinetic energy. On a rough slope that reads as: energy released by gravity, minus energy lost to friction, equals the kinetic energy gained. Written that way it handles problems that would need several equations of motion.

Energy accounting down a rough slope: 98 J released, 33.9 J lost to friction, 64.1 J left as kinetic energygravity 98 Jfriction 33.9 Jkinetic 64.1 Jwork done by gravity = friction loss + kinetic energyso v = √(2 × 64.05 / 5) = 5.06 m/s
FIG. 1Energy accounting down a rough slope: what gravity gives, what friction takes, and what is left as motion.

WORKED EXAMPLE

Down a rough slope

A particle of mass 5 kg slides 4 m from rest down a slope at 30° to the horizontal, with coefficient of friction 0.2. Find its speed at the bottom, taking g = 9.8 m/s².

Work by gravity = 5(9.8)(4)sin30° = 98 J.

Normal reaction = 5(9.8)cos30° = 42.4 N, so friction = 8.49 N and the work against it is 8.49 × 4 = 33.9 J.

Kinetic energy gained = 98 − 33.9 = 64.1 J, so ½(5)v² = 64.05 and v = 5.06 m/s.

Power

Power is the rate of doing work, measured in watts. For a vehicle moving at speed v against a driving force F:

P = Fv

So at a given power the driving force falls as the vehicle speeds up, which is why acceleration tails off. Maximum speed is reached when the driving force has dropped to equal the resistance, and setting P/v equal to the resistance solves for it in one line.

Power divided by speed gives the driving force; what is left after resistance accelerates the car1200 kg1333 N500 NP = Fv, so F = 20000 / 15resultant 833 N gives a = 0.694 m/s²
FIG. 2Engine power divided by speed giving the driving force, with the surplus over resistance producing the acceleration.

YOUR TURN

A car accelerating

A car of mass 1200 kg works at 20 kW. At the moment its speed is 15 m/s the resistance is 500 N. Find the acceleration, and the maximum speed if the resistance stays at 500 N.

Show the working

Driving force = 20000/15 = 1333 N.

Resultant = 1333 − 500 = 833 N, so a = 833/1200 = 0.694 m/s².

At maximum speed the acceleration is zero, so the driving force equals 500 N: 20000/v = 500 gives v = 40 m/s.

In practice resistance grows with speed, so the real maximum would be lower.

THE EXAM BIT

  • Include cos θ in the work done whenever the force is not along the motion.
  • Take the normal reaction out of energy equations: it never does work.
  • Set out the energy equation as a sentence of terms before substituting numbers.
  • For power questions, decide whether the vehicle is accelerating or at maximum speed before writing anything.

CHECK YOURSELF

A force of 30 N acts at 60° to the direction of motion while an object moves 8 m. Find the work done.

Show a hint

Only the component along the motion counts.

Show the answer

W

o

r

k

=

3

0

×

8

×

c

o

s

6

0

°

=

3

0

×

8

×

0

.

5

=

1

2

0

J

,

h

a

l

f

w

h

a

t

t

h

e

f

o

r

c

e

w

o

u

l

d

d

o

i

f

i

t

a

c

t

e

d

a

l

o

n

g

t

h

e

m

o

t

i

o

n

.

Work is Fd cos θ, and the work-energy principle says the total work done by all forces equals the change in kinetic energy.

Power is P = Fv, so the driving force falls as speed rises, and maximum speed is where it has dropped to equal the resistance.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.

  • Calculate work done by a force at an angle, and by gravity and friction.
  • Apply the work-energy principle to motion on a slope.
  • Use P = Fv to link engine power, driving force and acceleration.

Open the full revision checklist to see every objective in the course in one place.

No animated video for this topic yet; these notes stand alone.