The work done to move an object against a conservative force, like gravity, is equal to the change in its potential energy.
The gravitational potential energy ($U$) of a mass ($m$) at a distance ($r$) from the center of the Earth (mass $M$) is given by:
$U = -\frac{GMm}{r}$
Where $G$ is the gravitational constant.
$U_1 = -\frac{GMm}{R}$
$U_2 = -\frac{GMm}{2R}$
The work done ($W$) is the change in potential energy:
$W = \Delta U = U_2 - U_1$
$W = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right)$
$W = -\frac{GMm}{2R} + \frac{GMm}{R}$
$W = \frac{GMm}{R} \left(1 - \frac{1}{2}\right)$
$W = \frac{GMm}{2R}$
At the Earth's surface, the acceleration due to gravity is $g$. The force of gravity is $mg$. This force is also given by Newton's law of gravitation:
$mg = \frac{GMm}{R^2}$
From this, we can express $GM$ as:
$GM = gR^2$
Substitute $GM = gR^2$ into the work done equation:
$W = \frac{(gR^2)m}{2R}$
$W = \frac{mgR}{2}$
Therefore, the work done to raise the mass is $\frac{mgR}{2}$.
The power of a crane, which lifts a mass of $1000 \text{ kg}$ to a height of $20 \text{ m}$ in $10 \text{ s}$ is :
($g = 9.8 \text{ m/s}^2$)
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' (tangent in the plane) will be :
A frictionless circular wire of unit radius is fixed on the horizontal plane. Two point particles of unit mass start moving simultaneously from point $A \left( \theta = \frac{\pi}{2} \right)$ with identical uniform angular speeds in opposite directions, and meet again at point $B \left( \theta = -\frac{\pi}{2} \right)$. During this time, which of the following figures schematically represent the magnitude of the total linear momentum $\vec{P}$ of the system, as a function of $\theta$?

Suppose there is a uniform circular disc of mass $M \text{ kg}$ and radius $r \text{ m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is ________.
