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Home » Answer the following:(a) 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?(b) In the Figure, the man walks carrying a mass of on his hands. In Fig., he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of hangs at its other end. In which case is the work done greater?
Answer the following:
(a) 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?(b) In the Figure, the man walks 2 \mathbf{m} carrying a mass of 15 \mathrm{~kg} on his hands. In Fig., 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?
CBSE | Class 11 | NCERT | Physics | Work, Energy, and Power
Answer the following:
(a) 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?(b) In the Figure, the man walks 2 \mathbf{m} carrying a mass of 15 \mathrm{~kg} on his hands. In Fig., 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:

(a) As the satellite approaches the Earth, its potential energy drops, and since the system’s total energy should remain constant, the kinetic energy increases. As a result, the satellite’s velocity rises. Despite this, atmospheric friction reduces the total energy of the system by a small amount.

(b)

Scenario I:

\mathrm{m}=20 \mathrm{~kg}

Displacement of the object will be s=4 \mathrm{~m}

W=Fs\cos\theta

\theta is the angle between the force and displacement

\mathrm{Fs}=\mathrm{mgs} \cos \theta

\mathrm{W}=\mathrm{mgs} \cos \theta=20 \times 4 \times 9.8 \cos 90^{\circ}

W=0

\left(\cos 90^{\circ}=0\right)

Scenario II:

\begin{array}{l} \text { Mass }=20 \mathrm{~kg} \\ \mathrm{~S}=4 \mathrm{~m} \end{array}

The direction of the applied force and displacement is same.

\theta=0^{\circ}

\cos 0^{\circ}=1

\mathrm{W}=\mathrm{Fs} \cos \theta

=\operatorname{mgs} \theta

=20 \times 4 \times 9.8 \times \cos 0^{\circ}

=784 \mathrm{~J}

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

 

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