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:
$P= \frac{W}{t}$ (1)

The work done is equal to the elastic energy stored by the compressed spring:
$W=U= \frac{1}{2}kx^2$
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:
$W= \frac{1}{2}(5000 N/m)(0.04 m)^2=4 J$

And now we can use eq.(1) to calculate the average power output:
$P= \frac{W}{t}= \frac{4 J}{0.30 s} =13.3 W$