CAIE A-Level Mathematics 4.5 Energy, Work and Power Question Bank

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 = $1250 \times 36 \times 8$ ]
M1
For using Work Done = Force × Distance.
$\mathrm{WD}=360000 \mathrm{~J}$
A1
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 ]

$\mathrm{DF}=\frac{57000}{36}[=1583.3 .$ .
M1
Use changed Power in $P=\mathrm{DF} \times v$ .
$\frac{57000}{36}-1250=1400 a$
M1
For using Newton's 2nd law applied to the car.
$a=0.238 \mathrm{~ms}^{-2}$
A1

(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 ]

$\frac{64000}{32}=1250+1400 g \sin \theta$
M1
For using DF= resistance + component of the weight of
the car.
$\theta=3.1[3.0708 \ldots]$
A1

[Maximum number: 2]

Small smooth spheres A and B, of equal radii and of masses 5 kg and 3 kg respectively, lie on a smooth horizontal plane. Initially B is at rest and A is moving towards B with speed $8.5 \mathrm{~ms}^{-1}$ . The spheres collide and after the collision A continues to move in the same direction but with a quarter of the speed of B.

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

The kinetic energy before the collision is
$\frac12\times5\times8.5^2=180.625\text{ J}$ .
The kinetic energy after the collision is
$\frac12\times5\times2.5^2+\frac12\times3\times10^2=165.625\text{ J}$ .
Therefore, the loss of kinetic energy is
$180.625-165.625=15\text{ J}$ .

[Maximum number: 3]

Figure for Question 4.5 Energy, work and power question 3 — CAIE A-Level Mathematics AS

A machine for driving a nail into a block of wood causes a hammerhead to drop vertically onto the top of a nail. The mass of the hammerhead is 1.2 kg and the mass of the nail is 0.004 kg (see diagram). The hammerhead hits the nail with speed $v \mathrm{~m} \mathrm{~s}^{-1}$ and remains in contact with the nail after the impact. The combined hammerhead and nail move immediately after the impact with speed $40 \mathrm{~m} \mathrm{~s}^{-1}$ .

The nail is driven 4 cm into the wood.

Find the constant force resisting the motion.

$\begin{aligned} 0^{2}=(40)^{2}+2 \times 0.04 \times a \quad[a=-20000] \text { or } 0.04=\frac{0+40}{2} t \text { gets } t=0.002, \text { so } 0=40+0.002 a[a=-20000] \end{aligned}$

M1

Use of a 'suvat' method to get an equation in a. Allow sign errors. Allow $\pm 20000$ .

Do not allow 4 in place of 0.04 .

Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40 .

Attempt to use Newton's Second Law vertically.
$\begin{aligned} {[-R+(1.2+0.004) g=(1.2+0.004) \times a]} {[-R+12.04=1.204 a]} \end{aligned}$

M1

Must have the correct number of relevant terms. Allow sign errors, but terms including masses must be effectively added. Do not allow any mass other than (1.2
+ 0.004).

$R=24100 \mathrm{~N}\left[24092.04=\frac{602301}{25}\right]$

A1

WWW.

Note: use of wrong sign for g leads to answers 24067.96 which gets max M1M1A0.

Note: Missing weight term gets 24080 which gets Max M1M0A0.

3

2(b)

Alternative method for Question 2(b) using energy

[Change in PE =] $1.204 \mathrm{~g} \times 0.04[=0.4816]$
or [change in KE = ] $\frac{1}{2} \times 1.204 \times(40)^{2}[=963.2]$

B1

Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40 .
B0 for kinetic energy, if extra kinetic energy terms
present.

$1.204 \mathrm{~g} \times 0.04+\frac{1}{2} \times 1.204 \times(40)^{2}=0.04 R$

M1

Attempt at work energy equation. Must have correct number of relevant terms. dimensionally correct; allow sign errors.

Do not allow 4 in place of 0.04.

Allow use of 40.1 or $\frac{602}{15}$ for velocity in place of 40 .

$R=24100 \mathrm{~N}\left[24092.04=\frac{602301}{25}\right]$

A1

WWW

Note: use of wrong sign for g leads to answers 24067.96 which gets max B1M1A0.

Note: Missing potential energy term gets 24 080, which gets maximum of B1M0A0.

All question bank results loaded

Covered syllabus points

4.5 Energy, work and power question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

4.5 Energy, work and power question 2

4.5 Energy, work and power question 3