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)+1/2Kd^(2)`
C. `mg(h-d)-1/2Kd^(2)`
D. `mg(h+d)+1/2Kd^(2)`

Answer

A

Explanation

Loss in potential energy by ball on the spring
P.E = mg(h+d)

Spring acquired energy during compression by d
` U = 1/2Kd^(2)`
Work done by spring `rArr W = - U = (-1)/2 Kd^(2)`
Work done on teh system
The netwrok done = work done by ball + work done by spring
`W = mg(h+d)-1/2 Kd^(2)`
Alternative
Loss if potential energy of ball = mg `(h+d)`
Gain the potential energy by spring ` = -1/2 Kd^(2)`
Diring conservation of energy .
Loss P.E of ball is stored in teh form potential energy in spring so ,
`mg(h+d)=1/2 Kd^(2)`
But spring bal system is not at rest so the net work done by whole system
`W = mg(h+d) - 1/2Kd^(2)`