Question

A block of mass 0.250 kg is placed on top of a light, vertical spring of force constant 5 000 N/m and pushed downward so that the spring is compressed by 0.100 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?
Please confirm that I am doing this correctly...if not please provide detail.


Energy stored in spring is:
PE = ½·k·x²

So , PE = mgh
PE = ½·5000N/m·(0.100m)² = 25N·m
25N·m = 0.250kg·9.807m/s²·h
h = 10.197m

Answer 1

Answer:

The maximum height corresponds to 10.197m

Explanation:

Your approach is accurate because you are analyzing the energy changes and if this is conserved or not. In that perspective, the mass is located at in the top of a spring in which undergoes gravitational potential energy and when the mass compresses the spring, this has a restoring force given by $k=5000 \frac{N}{m}$ and governed by an elastic potential rnergy. Given the initial and final conditions, it can be described, as follows

$U_{gravitational}=U_{elastic}\\\\mgh=\frac{1}{2} k x^{2}\\ \\h_{max}=\frac{1}{2} \frac{k x^{2}}{mg}$

With the above analysis and considering the algebraic operations of the physical magnitudes, the maximum height  above the point of release is given by

$h_{max}=\frac{1}{2} \frac{k x^{2}}{mg}\\\\h_{max}=\frac{1}{2} \frac{5000 (N/m) (0.100 m)^{2}}{(0.250kg)(9.807 m/s^{2})}\\\\h=10.197m$