Question

2)

Answer 1

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

$x=0.150m-0.115m=0.035m$

And spring constant is $k=51.0N/m$

$PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312$

Hence, the potential energy is 0.0312J