Question

In just 0.30 s , you compress a spring (spring constant 5000 n/m ), which is initially at its equilibrium length, by 4.0 cm. part a what is your average power output?

Answer 1

The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:

`P= (W)/(t)` (1)

The work done is equal to the elastic energy stored by the compressed spring:

`W=U= (1)/(2)kx^2`
where
`k=5000 N/m` is the spring constant and
`x=4.0 cm=0.04 m` is the compression of the spring. If we substitute the numbers, we find:

`W= (1)/(2)(5000 N/m)(0.04 m)^2=4 J`

And now we can use eq.(1) to calculate the average power output:

`P= (W)/(t)= (4 J)/(0.30 s) =13.3 W`