Question

Answer the following:

(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

(c) An artificial satellite orbiting the earth in a very thin atmosphere loses its energy
gradually due to dissipation against atmospheric resistance, however small. Why
then does its speed increase progressively as it comes closer and closer to the earth?

d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig.
6.13(ii), he walks the same distance pulling the rope behind him. The rope goes
over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work
done greater?

Answer 1

Explanation:

a.)  When the casing burns up due to the friction, the rocket’s mass gets reduced.

As per the law of conservation of energy:

Total energy = kinetic energy + potential energy

= mgh + 1/2 mv^2

There will be a drop in total energy due to the reduction in the mass of the rocket. Hence, the energy which is needed for the burning of the casing is obtained from the rocket.

b.) The force due to gravity is a conservative force. The work done on a closed path by the conservative force is zero. Hence, for every complete orbit of the comet, the work done by the gravitational force is zero.

c.) The potential energy of the satellite revolving the Earth decreases as it approaches the Earth and since the system’s total energy should remain constant, the kinetic energy increases. Thus, the satellite’s velocity increases. In spite of this, the total energy of the system is reduced by a fraction due to the atmospheric friction.

d.) Ans.

Scenario I:

m = 20 kg

Displacement of the object, s = 4 m

W = Fs cos \thetaθ \thetaθ = It is the angle between the force and displacement

Fs = mgs cos \thetaθ

W = mgs cos \thetaθ = 20 x 4 x 9.8 cos 90∘

 = 0

 

Scenario II:

Mass = 20 kg

S = 4 m

The applied force direction is same as the direction of the displacement.

$\theta = 0^{\circ}$

 

Cos 0∘ = 1

W = Fs cos \thetaθ

= mgs \thetaθ

= 20 x 4 x 9.8 x cos 0∘

 

= 784 J

Thus, the work done is more in the second scenario.

Answer 2

Answer:

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