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(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 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. given below (i) the man walks $2 \mathrm{~m}$ carrying a mass of $15 \mathrm{~kg}$ in his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of $15 \mathrm{~kg}$ hangs at its other end. In which case is the work done greater?
(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 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. given below (i) the man walks $2 \mathrm{~m}$ carrying a mass of $15 \mathrm{~kg}$ in his hands. In Fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of $15 \mathrm{~kg}$ hangs at its other end. In which case is the work done greater?
Solution:
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(a) The heat energy required for burning of the casing of a rocket in flight is obtained from the rocket itself. The energy is obtained at the expense of the mass of the rocket and its kinetic and potential energy.
(b) The gravitational force on the comet due to Sun is a conservative force. Since the work done by a conservative force over a closed path is always zero (irrespective of the nature of path). Hence the work done by the gravitational force over every complete orbit of the comet is zero of the comet. .
(c) As the satellite comes closer and closer to Earth, its potential energy decreases. Since according to the law of conservation of energy, the sum of kinetic and potential energy must remain constant, so the kinetic energy and hence velocity of the satellite increases. However, the total energy of the satellite continuously decreases some-what due to the loss of energy against friction i.e. atmospheric resistance.
(d) In the case shown in fig. (i), the man applies the force on the $15 \mathrm{~kg}$ mass in vertically upward direction but walks $2 \mathrm{~m}$ in the horizontal direction so the angle between $F$ and $S$ is $90^{\circ}$, hence work done $W=F . S$ is zero.
In fig. (ii), the force applied on the mass of $15 \mathrm{~kg}$ in the vertically upward direction (= weight of the body) and the mass is also lifted along the vertical so $\theta=0$
$\therefore \mathrm{W}=m g S=15 \times 9.8 \times 2=294 \mathrm{~J}$
Thus work done in second case is greater.
(b) The gravitational force on the comet due to Sun is a conservative force. Since the work done by a conservative force over a closed path is always zero (irrespective of the nature of path). Hence the work done by the gravitational force over every complete orbit of the comet is zero of the comet. .
(c) As the satellite comes closer and closer to Earth, its potential energy decreases. Since according to the law of conservation of energy, the sum of kinetic and potential energy must remain constant, so the kinetic energy and hence velocity of the satellite increases. However, the total energy of the satellite continuously decreases some-what due to the loss of energy against friction i.e. atmospheric resistance.
(d) In the case shown in fig. (i), the man applies the force on the $15 \mathrm{~kg}$ mass in vertically upward direction but walks $2 \mathrm{~m}$ in the horizontal direction so the angle between $F$ and $S$ is $90^{\circ}$, hence work done $W=F . S$ is zero.
In fig. (ii), the force applied on the mass of $15 \mathrm{~kg}$ in the vertically upward direction (= weight of the body) and the mass is also lifted along the vertical so $\theta=0$
$\therefore \mathrm{W}=m g S=15 \times 9.8 \times 2=294 \mathrm{~J}$
Thus work done in second case is greater.
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