6.5 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 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 ?

0 5 Views | Posted 5 months ago
Asked by Shiksha User

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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    6.5 (a) As per the law of conservation of energy,

    Total energy = potential energy + kinetic energy

    = mgh + 12 mv2

    When the casing burns, mass reduces, resulting in drop of energy. Hence the energy for burning of casing is drawn from the rocket.

     

    (b) The force due to gravity is a conservative force. The work done on a closed path for a 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 k

    ...more

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(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force. 
 
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(b) conserving energy between “O” ans ”A”

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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-mgL + mgL=0

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

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(c) m =150g =3/20kg

Time of contact =0.001s

U=126km/h= 126 × 1000 60 × 60 = 35 m s

V= -35m/s

Change in momentum of the ball = m (v-u)= 3 20 - 35 - 35 k g m / s

=21/2

F= dp/dt=- 21 / 2 0.001 N = - 1.05 × 10 4 N

Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

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