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
