CAIE A-Level Mathematics AS 4.5 Energy Work and Power Questions

Practise calculating work, kinetic and potential energy and power, then applying work-energy or conservation equations to particles, vehicles and connected systems.

Exam points

calculate work as force times displacement in its direction, including signed resistance work
balance changes in kinetic and gravitational potential energy with work by external forces
use P = Fv for a driving force at instantaneous speed or work divided by elapsed time

Covered syllabus points

4.5.1—Scalar product 4.5.2—Gravitational potential energy 4.5.3—Forces 4.5.4—Complex loci 4.5.5—Instantaneous acceleration

[Maximum number: 9]

A car of mass 1400 kg is moving on a straight road against a constant force of 1250 N resisting the motion.

(a)

The car moves along a horizontal section of the road at a constant speed of $36 \mathrm{~m} \mathrm{~s}^{-1}$ .

[ 7 ]

(i)

Calculate the work done against the resisting force during the first 8 seconds.

[ 2 ]

[WD = $1250 \times 36 \times 8$ ]
For using Work Done = Force × Distance.
$\mathrm{WD}=360000 \mathrm{~J}$
or 360 kJ

(ii)

Calculate, in kW , the power developed by the engine of the car.

[ 2 ]

Power $=1250 \times 36$ or $\mathrm{P}=\frac{360000}{8}[=45000 \mathrm{~J}]$
B1 FT
FT Work Done from $\frac{\mathbf{a}(\mathbf{i})}{8}$ .
$=45 \mathrm{~kW}$
B1

(iii)

Given that this power is suddenly increased by 12 kW , find the instantaneous acceleration of the car.

[ 3 ]

$\mathrm{DF}=\frac{57000}{36}[=1583.3 .$ .
Use changed Power in $P=\mathrm{DF} \times v$ .
$\frac{57000}{36}-1250=1400 a$
For using Newton's 2nd law applied to the car.
$a=0.238 \mathrm{~ms}^{-2}$

(b)

The car now travels at a constant speed of $32 \mathrm{~m} \mathrm{~s}^{-1}$ up a section of the road inclined at $\theta^{\circ}$ to the horizontal, with the engine working at 64 kW .

Find the value of $\theta$ .

[ 2 ]

$\frac{64000}{32}=1250+1400 g \sin \theta$
For using DF= resistance + component of the weight of
the car.
$\theta=3.1[3.0708 \ldots]$

[Maximum number: 2]

Two particles A and B, of masses 3.2 kg and 2.4 kg respectively, lie on a smooth horizontal table. A moves towards B with a speed of $v \mathrm{~m} \mathrm{~s}^{-1}$ and collides with B, which is moving towards A with a speed of $6 \mathrm{~m} \mathrm{~s}^{-1}$ . In the collision the two particles come to rest.

Find the loss of kinetic energy of the system due to the collision.

$\mathrm{KE}= \pm \frac{1}{2} \times 3.2 \times(\text { their } 4.5)^{2}$ OR $\pm \frac{1}{2} \times 2.4 \times 6^{2}$
Attempt at either KE term, using their v.

Marking guidance:

Do not allow $\frac{1}{2} \times 3.2 \times(\text { their } 4.5 \pm 6)^{2}$ ,
or $\frac{1}{2} \times 2.4 \times(\text { their } 4.5 \pm 6)^{2}$ ,
or $\frac{1}{2} \times(3.2+2.4) \times(\text { their } 4.5 \pm 6)^{2}$ ,
or $\frac{1}{2} \times 3.2 \times(\text { their } 4.5-0)^{2}$ ,
or $\frac{1}{2} \times 2.4 \times(6-0)^{2}$ .

$\mathrm{KE}_{\text {loss }}=75.6 \mathrm{~J}$
Allow -75.6.

Note $\frac{1}{2} \times(3.2+2.4) \times 6^{2}$ or $\frac{1}{2} \times(3.2+2.4) \times(\text { their } 4.5)^{2}$ is M1A0.

[Maximum number: 11]

A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is 80 kg . There is a constant resistance force of magnitude 32 N to the cyclist's motion. At an instant when she is travelling at $7 \mathrm{~ms}^{-1}$ , her acceleration is $0.1 \mathrm{~ms}^{-2}$ .

(a)

Find the power output of the cyclist.

[ 3 ]

$D F=\frac{P}{7}$
For $P=D F \times 7$ OE seen at any point in working.

Marking guidance:

Allow any force term or simply D F,
e.g. $32,80 \times 0.1,32-80 \times 0.1,80 \times 10$ etc.

$D-32=80 \times 0.1$
For use of Newton's second law. Must have correct number of
terms. Allow sign errors.

[Power = ] 280 W

(b)

Find the steady speed that the cyclist can maintain if her power output and the resistance force are both unchanged.

[ 2 ]

[At steady speed driving force =] $32=\frac{280}{v}$

M1

Attempt at equilibrium equation ( a=0 ) with their power.

Steady speed $=8.75 \mathrm{~ms}^{-1}$ or $\frac{35}{4} \mathrm{~ms}^{-1}$

A1FT

OE

FT their power from part (a) $\frac{\text { their } 280}{32}$ .

(c)

The cyclist later descends a straight hill of length 32.2 m , inclined at an angle of $\sin ^{-1}\left(\frac{1}{20}\right)$ to the horizontal. Her power output is now 120 W , and the resistance force now has variable magnitude such that the work done against this force in descending the hill is 1128 J . The time taken to descend the hill is 4 s .

Given that the speed of the cyclist at the top of the hill is $7.5 \mathrm{~m} \mathrm{~s}^{-1}$ , find her speed at the bottom of the hill.

[ 6 ]

$120 \times 4 \quad[480]$
Work done by cyclist.

$\frac{1}{2} \times 80 v^{2}$ $\left[40 v^{2}\right]$

or $\frac{1}{2} \times 80 \times 7.5^{2}$

[2250]
For at least one KE term.

$80 g \times \frac{1}{20} \times 32.2 \quad[1288]$ $[80 \times 10 \times 1.61]$
Change in PE.

Attempt at work-energy equation
Attempt at work energy equation with five relevant terms (four relevant terms plus work done against resistance);
dimensionally
correct. Allow sign errors. Allow sin/cos mix.

$120 \times 4+80 g \times \frac{1}{20} \times 32.2-1128=\frac{1}{2} \times 80\left(v^{2}-7.5^{2}\right)$ $\left[480+1288-1128=40 v^{2}-2250\right]$
For correct equation.

Speed $=8.5[0] \mathrm{m} \mathrm{s}^{-1}$ or $\frac{17}{2}$
OE

Use of constant acceleration scores M0 and cannot score B marks if the method leading to their answer only uses constant acceleration.

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Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question 2

Question 3

Question (a)

Question (b)

Question (c)