• Physics 9702 Doubts | Help Page 233

Question 1094: [Work, Energy and Power]

A constant force of 9.0 kN, parallel to an inclined plane, moves a body of weight 20 kN through a distance of 40 m along the plane at constant speed. The body gains 12 m in height, as shown.

How much of the work done is dissipated as heat?

A 120 kJ                      B 240 kJ                      C 360 kJ                      D 600 kJ

Reference: Past Exam Paper – November 2004 Paper 1 Q18

Solution 1094:

Answer: A.

The work done by the constant force of 9.0 kN to move the body a distance of 40 m is

Work done = 9.0 × 40 = 360 kJ

This energy is converted to the gravitational potential energy of the body. Some of the energy is also dissipated as heat.

Gain in gravitational potential energy of the body = 20 × 12 = 240 kJ

Work done = Gain in GPE + Heat dissipated

Heat dissipated = 360 – 240 = 120 kJ

Question 1095: [Circular motion]

(a) (i) Define the radian.

(ii) A small mass is attached to a string. The mass is rotating about a fixed point P at constant speed, as shown in Fig. 1.1.

Fig. 1.1

Explain what is meant by the angular speed about point P of the mass.

(b) A horizontal flat plate is free to rotate about a vertical axis through its centre, as shown in Fig. 1.2.

Fig. 1.2

A small mass M is placed on the plate, a distance d from the axis of rotation.

The speed of rotation of the plate is gradually increased from zero until the mass is seen to slide off the plate.

The maximum frictional force F between the plate and the mass is given by the expression

F = 0.72W,

where W is the weight of the mass M.

The distance d is 35 cm.

Determine the maximum number of revolutions of the plate per minute for the mass M to remain on the plate. Explain your working.

(c) The plate in (b) is covered, when stationary, with mud.

Suggest and explain whether mud near the edge of the plate or near the centre will first leave the plate as the angular speed of the plate is slowly increased.

Reference: Past Exam Paper – June 2008 Paper 4 Q1

Solution 1095:

(a) (i) The radian is defined as the angle (subtended) at centre of a circle by an arc equal in length to the radius (of the circle)

(ii) The angular speed is the angle swept out per unit time / rate of change of angle by the string

(b)

The friction {between the mass and the plate} provides / equals the centripetal force

{Centripetal force = mrω²                      where r = d = 35 cm = 0.35 m}

0.72 W = mdω²

0.72 mg = m × 0.35ω²

Angular speed ω = 4.49 (rad s^–1)

{The angular speed is the angle swept out per unit time. For 1 revolution, the angle swept is 2π.

But the angular speed ω = 4.49 rad s^–1. That is, in 1 second, an angle of (ω =) 4.49 rad is swept. In 1 minute (= 60 seconds) the angle swept is (4.49 × 60) rad.

For n revolutions, the angle swept is n(2π).

In 1 minute (= 60 seconds) the angle swept is (4.49 × 60) rad.

Number of revolutions in 1 minute = (4.49 × 60) / 2π }

n = (ω /2π) × 60

n = 43 min^–1  

(c) EITHER The centripetal force increases as r increases     OR centripetal force larger at the edge, so the mud flies off at edge first

Question 1096: [Simple harmonic motion]

(a) The defining equation for simple harmonic motion is

a = −ω²x.

For November 2003: (i) Identify the symbols in the equation.

For Specimen Papers: (i) State the relation between ω and the frequency f.

(ii) State the significance of the negative (−) sign in the equation.

(b) A frictionless trolley of mass m is held on a horizontal surface by means of two similar springs, each of spring constant k. The springs are attached to fixed points, as illustrated in Fig. 3.1.

Fig. 3.1

When the trolley is in equilibrium, the extension of each spring is e.

The trolley is then displaced a small distance x to the right along the axis of the springs. Both springs remain extended.

(i) Show that the magnitude F of the restoring force acting on the trolley is given by

F = 2kx.

(ii) The trolley is then released. Show that the acceleration a of the trolley is given by

a = − 2kx / m.

(iii) The mass m of the trolley is 900 g and the spring constant k is 120 N m⁻¹. By comparing the equations in (a) and (b)(ii), determine the frequency of oscillation of the trolley.

For November 2003:

(c) Suggest why the trolley in (b) provides a simple model for the motion of an atom in a crystal.

Reference: Past Exam Paper – November 2003 Paper 4 Q2 & Specimen 2004 Paper 4 Q2 & Specimen 2016 Paper 4 Q3

Solution 1096:

(a)

For November 2003: a: acceleration, ω: angular frequency and x: displacement

For Specimen Papers: ω = 2π f

(ii)

EITHER There is a (−)ve because a and x are in opposite directions

OR a is always directed towards mean position

(b)

(i)

{From Hooke’s law: Force F = (-) k × extension

Initially, the extension of each spring is e. The trolley is then displaced a small distance x to the right. This causes the extension of the spring on the left to increase by x (becoming e+x) and the extension of the spring on the right to decrease by x (becoming e-x).}

Forces in springs are k(e + x) and k(e – x)

{The forces due to each spring opposes each other. So we should consider the difference between them.}

Resultant = k(e + x) − k(e – x)

Resultant = 2kx

(ii)

Resultant force F = ma

{Resultant (restoring) force = 2kx}

Acceleration a = −2kx / m

The (−) sign accounts for the fact that the acceleration is always opposite to the displacement.

(iii)

{From (a): a = −ω²x                            From (b)(ii): a = −2kx / m

Thus, ω² = 2k / m}

ω² = 2k / m

(2πf)² = (2 × 120) / 0.90

Frequency f = 2.6 Hz

For November 2003:

(c) Choose any 2:

The atom is held in position by attractive forces

The atom oscillates,

There is not just two forces     OR it is 3D not 1D

The force is not proportional to x