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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
(1) Fig. 6.13
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 very thin atmosphere loses its energy
15 kg
15 kg
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 ?
Answers
Answer:
ask these questions separately
Answer:
Heat energy required for burning of casing of rocket comes from the rocket itself. As a result of work done against friction the kinetic energy of rocket continuously deceases and this work against friction reappears as heat energy.
Gravitational force is a conservative force. Hence the work done depends on the initial and final positions.
2.In a complete revolution of the sun by a comet, the displacement becomes zero, so the work done by the gravitational force over every complete orbit of the comet is zero.
3.As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential decreases rapidly. As a result, the kinetic energy of satellite slightly increases i.e. its speed increases progressively.
4.In the second case Case i Mass m = 15 kg Displacement s = 2 m Case ii Mass m = 15 kg Displacement s = 2 mHere the direction of the force appliedon the rope and the direction of the displacement of the rope is same.Therefore the angle between them θ = 0°Since cos 0° = 1Work done W = Fs cosθ = mgs= 15 x 9.8 x 2 = 294 JHence more work is done in the second case.