## Wednesday, September 06, 2006

### Work Done in Lifting a Body

When you lift a body of mass ‘m’ from the earth’s surface through a small height ‘h’, the work ‘W’ done by you against gravitational force, as you know, is given by W = mgh. This expression gives you the correct value only if ‘h’ is negligibly small compared to the radius ‘R’ of the earth.
If the height ‘h’ is comparable to the radius ‘R’ of the earth, the expression gets modified as W = mgh/[1+(h/R)].
If h = nR the above expression can be written as W = mgRn/(n+1).
In all these expressions, ‘g’ is the surface value of acceleration due to gravity.
Let us consider the following multiple choice questions:
(1)The work done in lifting a body of mass ‘m’ from the earth’s surface, through a height ‘h’ is mgR/3 where ‘g’ is the acceleration due to gravity on the earth’s surface and ‘R’ is the radius of the earth. Then ‘h’ is equal to
(a) R (b) R/2 (c) R/3 (d) R/4 (e) R/6

We have W = mgh/[1+(h/R)] so that mgR/3 = mgh/[1+(h/R)]. This gives h = R/2 [option (b)].
(2) A body of mass ‘m’ is projected at an angle of 45˚ from the moon’s surface by giving it kinetic energy equal to mgR where ‘R’ is the radius of the moon and ‘g’ is the acceleration due to gravity on the moon’s surface. Then
(a)the final potential energy of the body will be zero (b) the final kinetic energy of the body will be positive (c) the body will reach a height equal R (d) the body will reach a height between R and 2R (e) the body will reach a height between 2R and 4R.
The potential energy of the body on the moon’s surface is –GMm/R = -mgR, on substituting g = GM/R2. When the body is given kinetic energy equal to mgR, its total energy becomes zero and it escapes into the outer space where its potential energy and kinetic energy are zero. The correct option therefore is (a).
You can find more multiple choice questions (with solution) of similar type here