Question

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

Answer 1

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant $k=4.45\times 10^3 N/m$

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

$P_i=\frac{kx^2}{2}$

$P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J$

$P_f=mgh=0.25\times 9.8\times h$

$P_i=P_f$

$14.24 =0.25\times 9.8\times h$

$h=\frac{14.24}{0.25\times 9.8}=5.81 m$