Mathematics 9709/43 — May/June 2016

Cambridge A-Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Kinematics of Motion in a Straight Line · Forces and Equilibrium · Energy, Work and Power · Newton's Laws of Motion

Q1Energy, Work and PowerFree sample

A particle of mass $8\text{ kg}$ is pulled at a constant speed a distance of $20\text{ m}$ up a rough plane inclined at an angle of $30^{\circ}$ to the horizontal by a force acting along a line of greatest slope.

(i)

Find the change in gravitational potential energy of the particle.

DifficultyMedium-Easy

Worked solution

Approach

The change in gravitational potential energy equals the work done against gravity. The vertical height gained is the distance along the plane multiplied by the sine of the angle of inclination.

Working

The vertical height gained is:

h = 20\sin 30^{\circ} = 20 \times \frac{1}{2} = 10\text{ m}

Using $g = 10\text{ m s}^{-2}$ (as implied by the mark scheme), the change in gravitational potential energy is:

\Delta \text{PE} = mgh = 8 \times 10 \times 10 = 800\text{ J}

Answer

800\text{ J}

Final answer

800 J

Detailed explanation

Walkthrough

We need the change in gravitational potential energy. The particle moves 20 m up the plane, which is inclined at $30^{\circ}$ . The vertical height gained is the component of the displacement in the vertical direction. Since the plane makes an angle of $30^{\circ}$ with the horizontal, the vertical rise is $20\sin 30^{\circ} = 10$ m. Then use $\Delta \text{PE} = mgh$ with $m = 8$ kg and $g = 10$ m/s² (the value used by the mark scheme) to get 800 J.

Key Takeaways

Gravitational potential energy depends only on the vertical height gained, not on the path taken. For motion along an incline, the vertical height is $d\sin \theta$ . The change in PE is $mgh$ .

Common Mistakes

Using the distance along the plane (20 m) as the height instead of $20\sin 30^{\circ}$ .
Using $\cos 30^{\circ}$ instead of $\sin 30^{\circ}$ .
Using $g = 9.8$ and getting 784 J, which is not the mark scheme answer (they use $g = 10$ ).

Things to Be Careful About

The value of $g$ : the mark scheme implies $g = 10$ m/s². If the question specifies a value, use that.
Ensure the height is in metres and mass in kg to get energy in joules.

Techniques used

compute vertical height using trigonometryapply gravitational potential energy formula

(ii)

The total work done against gravity and friction is $1146\text{ J}$ . Find the frictional force acting on the particle.

DifficultyMedium-Easy

Worked solution

Approach

The total work done against gravity and friction is given. The work done against gravity is the change in gravitational potential energy from part (i). Subtract this from the total to find the work done against friction, then divide by the distance to find the frictional force.

Working

Work done against gravity = 800 J.

Total work = 1146 J, so work against friction:

W_{\text{friction}} = 1146 - 800 = 346\text{ J}

Since the frictional force acts along the plane opposite to the motion, the work done against friction is $W = Fd$ , where $d = 20\text{ m}$ :

F = \frac{346}{20} = 17.3\text{ N}

Answer

17.3\text{ N}

Final answer

17.3 N

Detailed explanation

Walkthrough

We are told the total work done against gravity and friction is 1146 J. From part (i), the work done against gravity (which equals the change in PE) is 800 J. Therefore the work done against friction is $1146 - 800 = 346$ J. The frictional force acts along the plane opposite to the direction of motion, so the work done against friction is $F \times d$ , where $d = 20$ m. Hence $F = \frac{346}{20} = 17.3$ N.

Key Takeaways

The total work done against all resistive forces equals the sum of the work done against each force. Work done by a constant force along the direction of motion is $Fd$ . When a force opposes motion, the work done against it is still $Fd$ (magnitude).

Common Mistakes

Forgetting to subtract the work against gravity (800 J) from the total, and dividing 1146 by 20 directly.
Using the total work as the work against friction.
Forgetting to divide by the distance to get the force.

Things to Be Careful About

The frictional force is constant, so $W = Fd$ applies.
Units: work in joules, distance in metres, force in newtons.
The angle between the frictional force and displacement is $180^{\circ}$ , so the work done by friction is negative, but we are asked for the work done against friction, which is positive.

Techniques used

subtract work against gravity from total workdivide work against friction by distance to find force

The rest of this paper

6 more questions

Q2Kinematics of Motion in a Straight Line5M
Q3Forces and Equilibrium6M
Q4Forces and Equilibrium7M
Q5Energy, Work and Power · Forces and Equilibrium · Newton's Laws of Motion8M
Q6Newton's Laws of Motion · Kinematics of Motion in a Straight Line10M
Q7Kinematics of Motion in a Straight Line10M

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