Work Done Against Gravity MCQ - Practice Questions & Answers

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If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is :

$2mgR$

$\frac{1}{2}mgR$

$\frac{1}{4}mgR$

$mgR$

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Concepts Covered - 1

Work Done Against Gravity

The gravitational potential energy at height 'h' from the earth's surface

Is given by $U_{h}=-\frac{mgR}{1+\frac{h}{R}}$

So at the surface of earth put h=0

We get $U_s=-mgR$

So if the body of mass m is moved from the surface of earth to a point at height h from the earth's surface

Then there is a change in its potential energy.

And this change in its potential energy is known as work done against gravity to move the body from earth surface to height h.

$W=\Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]$

Where $W \rightarrow$ work done

$\Delta U \rightarrow$ change in Potential energy

$r_{1},r_{2}\rightarrow$ distances

Putting r₁=R , and r₂=R+h

So $W=\Delta U=GMm\left [ \frac{1}{R}-\frac{1}{R+h} \right ]$

when 'h' is not negligible

$W=\frac{mgh}{1+\frac{h}{R}}$

when 'h' is very small

As $W=\frac{mgh}{1+\frac{h}{R}}$

But $h$ is small as compared to earth's radius

$\frac{h}{R}\rightarrow 0$

So $W=mgh$

If h = nR then

$W=mgR*\frac{n}{n+1}$

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Work Done Against Gravity - Practice Questions & MCQ

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