Exam-Style Problems

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June 2012 p41 q5

3558

The diagram shows the vertical cross-section OAB of a slide. The straight line AB is tangential to the curve OA at A. The line AB is inclined at \(\alpha\) to the horizontal, where \(\sin \alpha = 0.28\). The point O is 10 m higher than B, and AB has length 10 m (see diagram). The part of the slide containing the curve OA is smooth and the part containing AB is rough. A particle P of mass 2 kg is released from rest at O and moves down the slide.

Find the speed of P when it passes through A.
The coefficient of friction between P and the part of the slide containing AB is \(\frac{1}{12}\). Find
1. the acceleration of P when it is moving from A to B

Solution

(i) The potential energy (PE) loss from O to A is given by:

\(\text{PE loss} = 2g(10 - 10 \times 0.28)\)

Using conservation of energy, \(\frac{1}{2} mv^2 = \text{PE loss}\):

\(\frac{1}{2} \times 2 \times v^2 = 144\)

Solving for \(v\), we find \(v = 12 \text{ m/s}\).

(ii) The normal reaction \(R\) is given by:

\(R = 2g \times 0.96\)

Using Newton's second law, the equation of motion is:

\(2g \times 0.28 - 2g \times 0.96 \times \frac{1}{12} = 2a\)

Solving for \(a\), we find \(a = 2 \text{ m/s}^2\).

(iii) Using the equation \(v^2 = u^2 + 2as\) from A to B:

\(v^2 = 12^2 + 2 \times 2 \times 10\)

Solving for \(v\), we find \(v = 13.6 \text{ m/s}\).

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Nov 2022 p43 q2

3559

A box of mass 5 kg is pulled at a constant speed of 1.8 m/s for 15 s up a rough plane inclined at an angle of 20° to the horizontal. The box moves along a line of greatest slope against a frictional force of 40 N. The force pulling the box is parallel to the line of greatest slope.

(a) Find the change in gravitational potential energy of the box.

(b) Find the work done by the pulling force.

Solution

(a) The change in gravitational potential energy (PE) is given by:

\(\text{PE} = mgh\)

where \(m = 5 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), \(h = s \sin \theta\), and \(s = vt = 1.8 \times 15 \text{ m}\).

\(h = 1.8 \times 15 \times \sin 20^{\circ}\)

\(\text{PE} = 5 \times 9.8 \times 1.8 \times 15 \times \sin 20^{\circ}\)

\(\text{PE} = 462 \text{ J}\)

(b) The work done (WD) by the pulling force is the sum of the work done against gravity and the work done against friction:

\(\text{WD} = \text{PE} + F_f \times s\)

\(\text{WD} = 5 \times 9.8 \times 1.8 \times 15 \times \sin 20^{\circ} + 40 \times 1.8 \times 15\)

\(\text{WD} = 1540 \text{ J}\)

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June 2012 p41 q3

3560

A load of mass 160 kg is pulled vertically upwards, from rest at a fixed point O on the ground, using a winding drum. The load passes through a point A, 20 m above O, with a speed of 1.25 m s^-1 (see diagram). Find, for the motion from O to A,

the gain in the potential energy of the load,

the gain in the kinetic energy of the load.

The power output of the winding drum is constant while the load is in motion.

Given that the work done against the resistance to motion from O to A is 20 kJ and that the time taken for the load to travel from O to A is 41.7 s, find the power output of the winding drum.

Solution

(i) The gain in potential energy (PE) is given by the formula:

\(\Delta PE = mgh\)

where \(m = 160 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 20 \text{ m}\).

\(\Delta PE = 160 \times 9.8 \times 20 = 32,000 \text{ J}\)

(ii) The gain in kinetic energy (KE) is given by the formula:

\(\Delta KE = \frac{1}{2} mv^2\)

where \(m = 160 \text{ kg}\) and \(v = 1.25 \text{ m/s}\).

\(\Delta KE = \frac{1}{2} \times 160 \times (1.25)^2 = 125 \text{ J}\)

(iii) The total work done (WD) by the drum is the sum of the gain in potential energy, gain in kinetic energy, and the work done against resistance:

\(WD = 32,000 + 125 + 20,000 = 52,125 \text{ J}\)

The power output \(P\) is given by:

\(P = \frac{\Delta (WD)}{\Delta T}\)

where \(\Delta T = 41.7 \text{ s}\).

\(P = \frac{52,125}{41.7} = 1250 \text{ W}\)

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Nov 2011 p43 q4

3561

ABC is a vertical cross-section of a surface. The part of the surface containing AB is smooth and A is 4 m higher than B. The part of the surface containing BC is horizontal and the distance BC is 5 m (see diagram). A particle of mass 0.8 kg is released from rest at A and slides along ABC. Find the speed of the particle at C in each of the following cases.

The horizontal part of the surface is smooth.

The coefficient of friction between the particle and the horizontal part of the surface is 0.3.

Solution

(i) The potential energy (PE) at A is given by \(0.8g \times 4\), where \(g\) is the acceleration due to gravity. This energy is converted into kinetic energy (KE) at C. Therefore, \(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 4\). Solving for \(v\), we get \(v = 8.94 \text{ ms}^{-1}\).

(ii) The force of friction \(F\) is \(0.3 \times 0.8g\). The work done against friction over the distance BC is \(F \times 5\). The loss in potential energy is \(0.8g \times 4\). The kinetic energy at C is given by \(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 4 - F \times 5\). Solving for \(v\), we get \(v = 7.07 \text{ ms}^{-1}\).

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Nov 2011 p42 q6

3562

A lorry of mass 16000 kg climbs a straight hill ABCD which makes an angle \(\theta\) with the horizontal, where \(\sin \theta = \frac{1}{20}\). For the motion from A to B, the work done by the driving force of the lorry is 1200 kJ and the resistance to motion is constant and equal to 1240 N. The speed of the lorry is 15 m/s at A and 12 m/s at B.

Find the distance AB.

For the motion from B to D the gain in potential energy of the lorry is 2400 kJ.

Find the distance BD.

For the motion from B to D the driving force of the lorry is constant and equal to 7200 N. From B to C the resistance to motion is constant and equal to 1240 N and from C to D the resistance to motion is constant and equal to 1860 N.

Given that the speed of the lorry at D is 7 m/s, find the distance BC.

Solution

(i) The kinetic energy loss is given by:

\(\frac{1}{2} \times 16000 \times (15^2 - 12^2)\)

The potential energy gain is:

\(16000g \times \frac{AB}{20}\)

Using the work-energy principle:

\(1200 = 0.8g(AB) + 1.24(AB) - 648\)

Solving for \(AB\), we find:

Distance \(AB = 200 \text{ m}\)

(ii) The gain in potential energy from \(B\) to \(D\) is 2400 kJ, which corresponds to:

\(2400 = 16000g \times \frac{BD}{20}\)

Solving for \(BD\), we find:

Distance \(BD = 300 \text{ m}\)

(iii) Work done against resistance is:

\(1240(BC) + 1860(300 - BC)\)

Using the work-energy principle:

\(\frac{1}{2} \times 16000 \times (12^2 - 7^2) = 2400000 + (558000 - 620BC) - 7200 \times 300\)

Solving for \(BC\), we find:

Distance \(BC = 61.3 \text{ m}\)

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