Edexcel A-Level Mathematics A2 M2.3 Work and Energy Questions

Practise M2 work and energy by linking kinetic energy, gravitational potential energy, resistance, friction and power in particle or vehicle models.

Exam points

Build a work-energy equation with KE change, GPE change and work against friction or resistance.
Use P=Fv or conservation of mechanical energy to find speed, acceleration or total work done.

Covered syllabus points

M2.3.1 - Kinetic and potential energy, work

[Maximum number: 3]

Figure 5

A thin smooth hollow spherical shell has centre O and radius r. Part of the shell is removed to form a bowl with a plane circular rim. The bowl is fixed with the circular rim uppermost and horizontal. The point A is the lowest point of the bowl, as shown in Figure 5.

The point B is on the rim of the bowl, with OB at an angle $\theta$ to the upward vertical, where $\tan \theta=\frac{12}{5}$
A small ball is placed in the bowl at A. The ball is projected from A with horizontal speed u and moves in the vertical plane AOB. The ball stays in contact with the bowl until it reaches B.

At the instant when the ball reaches B, the speed of the ball is v.
By modelling the ball as a particle and ignoring air resistance,

use the principle of conservation of mechanical energy to show that

v^{2}=u^{2}-\frac{36}{13} g r

Use conservation of energy equation with 2KE terms and 1EPE term. Note there are rearrangements. Dimensionally correct, terms of the correct structure,
condone sign errors. EPE of the form $\frac{1}{2} k x^{2}$
For an unsimplified equation with at most one error
For a correct unsimplified equation

A1*

Given answer obtained from complete and correct working. Must include a line
of working before reaching the given answer.

[Maximum number: 13]

A cyclist is travelling on a straight horizontal road and working at a constant rate of 500 W .

The total mass of the cyclist and her cycle is 80 kg .
The total resistance to the motion of the cyclist is modelled as a constant force of magnitude 60 N .

(a)

Using this model, find the acceleration of the cyclist at the instant when her speed is $6 \mathrm{~m} \mathrm{~s}^{-1}$

On the following day, the cyclist travels up a straight road from a point A to a point B.
The distance from A to B is 20 km.
Point A is 500 m above sea level and point B is 800 m above sea level.
The cyclist starts from rest at A.
At the instant she reaches B her speed is $8 \mathrm{~ms}^{-1}$
The total resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude 60 N.

[ 4 ]

Use P=Fv:

$F=\frac{500}{6}$
Equation of motion:

F-60=80a $a=\frac{7}{24}\ \mathrm{m\,s^{-2}}$

(b)

Using this model, find the total work done by the cyclist in the journey from A to B.

Later on, the cyclist is travelling up a straight road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha=\frac{1}{20}$

The cyclist is now working at a constant rate of P watts and has a constant speed of $7 \mathrm{~m} \mathrm{~s}^{-1}$

The total resistance to the motion of the cyclist from non-gravitational forces is again modelled as a constant force of magnitude 60 N.

[ 5 ]

$\text{Gain in KE}=\frac12\times80\times8^2=2560\ \mathrm J$ $\text{Gain in GPE}=80\times9.8\times300=235200\ \mathrm J$ $\text{Work against resistance}=20000\times60$ $\text{Total work}=40\times64+80\times9.8\times300+20000\times60=1437760\ \mathrm J$ $1440\ \mathrm{kJ}\quad\text{or}\quad 1400\ \mathrm{kJ}$

(c)

Using this model, find the value of P

[ 4 ]

Equation of motion
$\begin{aligned} F-60-80g\sin\alpha&=0\\ \frac{P}{7}-60-80g\left(\frac1{20}\right)&=0 \end{aligned}$ $P=694\ \text{W}\quad\text{or}\quad 690\ \text{W}$

[Maximum number: 8]

A rough plane is inclined to the horizontal at an angle $\alpha$ , where $\tan \alpha=\frac{3}{4}$

A particle P of mass m is held at rest at a point A on the plane.
The particle is then projected with speed u up a line of greatest slope of the plane and comes to instantaneous rest at the point B.

The coefficient of friction between the particle and the plane is $\frac{1}{7}$

(a)

Given that $u=\sqrt{10ag}$ , use the work-energy principle to find AB in terms of a.

[ 4 ]

Energy equation: PE gain plus work done against friction equals KE lost, or equivalent.
The question tells candidates to use work-energy. Need all terms and dimensionally correct;
condone sign errors and sine/cosine confusion.
$\frac{4mgd}{35}+mgd\sin\alpha=\frac12m\times10ag$
or
$\frac{4mgd}{35}+mgd\times\frac35=\frac12m\times10ag$
Unsimplified equation with at most one error;
correct unsimplified equation.
d=AB=7a
cao

(b)

to find, in terms of a and g, the speed of P when it returns to A.

[ 4 ]

Energy equation.
Need all terms and dimensionally correct;
condone sign errors.
$\frac{4mg}{35}\times14a=\frac12m\times10ag-\frac12mV^2$
or
$\frac{4mg}{35}\times7a=mg\times7a\times\frac35-\frac12mV^2$
A1ft
A1ft
$V=\sqrt{\frac{34ag}{5}}$
Accept $2.6\sqrt{ag}$ , $\sqrt{6.8ag}$ , or better.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)