A vertical spring with force constant `k` is fixed on a table. A ball of mass `m` at a height `h` above the free upper end of the spring falls vertically on the spring , so that the spring is compressed by a distance `d`. The net work done in the process is

Options

A. `mg(h+d)- (1)/(2) kd^(2)`
B. `mg (h-d)-(1)/(2) kd^(2)`
C. `mg (h-d)+(1)/(2) kd^(2)`
D. `mg (h+d)+(1)/(2) kd^(2)`.

Answer

A

Explanation

When a mass falls on a spring from a height h the work done by the loss of potential energy of the mass is stored as the potential energy of the spring.
One can write from conservation of energy that, `mg(h+d) = (1)/(2) kd^(2)`

`mg(h+d) = (1)/(2) kx^(2) = (1)/(2) kd^(2)`
The two energies are equal.
If work done is initial P.E. - final P.E., it is zero.
Work done is totally converted (assuming there is no loss). The work done in compression or expansion is always positive as it is directly proportional to `x^2`. The answer expected is `mg(h+d) - (1)/(2) kd^(2)` or, `(1)/(2) kd^(2) - mg(h+d)` as seen from options, but it is not justified. Question could have been more specific like work done by oscillation.