Question

A block of mass 0.254 kg is placed on top of a light, vertical spring of force constant 5 100 N/m and pushed downward so that the spring is compressed by 0.093 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)

Answer 1

Answer:

8.86 m

Explanation:

According to the law of conservation of energy, the elastic potential energy initially stored in the spring will be converted into gravitational potential energy of the block when it is at its maximum height:

$\frac{1}{2}kx^2 = mgh$

where

k = 5100 N/m is the spring constant

x = 0.093 m is the spring compression

m = 0.254 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

h is the maximum height of the block

Solving the equation for h, we find

$h=\frac{kx^2}{2mg}=\frac{(5100 N/m)(0.093 m)^2}{2(0.254 kg)(9.8 m/s^2)}=8.86 m$