Q3 / 92 · NTA Abhyas (JEE Main) 2020 · Physics · Work Power Energy

A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is

Correct answer
A
$\frac{m g R}{2}$
B
$m g R$
C
$2 m g R$
D
$\frac{m g R}{4}$

Solution

Work done

= ∆ P E = - \frac{G M m}{R + R} - (- \frac{G M m}{R})

= \frac{G M m}{2 R} = \frac{m g R}{2}

Hence, the correct answer is option: 'A'.

More Work Power Energy Questions

An object of mass 8 kg starts to move from rest under the action of a variable force F = 3 x N where x is the distance (in m ) covered by th
The system is released form rest with both the springs in unstretched position. Mass of each block is 1 kg and force constant of each spring
A chain (mass M ) is hanging from a wooden structure as shown in the figure A . It is slowly taken to a state as shown in the figure B . Cal
A block of mass m is gently placed over a massive plank moving horizontally over a smooth surface with velocity 10 m s - 1 . The coefficient
A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m s - 1 . It then ascends a smooth continuous trac
Power applied to a particle varies with time as P = ( 3 t 2 - 2 t + 1 ) watt, where t is in second. What is the change in its kinetic energy