Question

A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answer to two significant figures.

Answer 1

The answer is 4800 on the assignment. good luck

Answer 2

Answer:

The spring constant of the spring is $k=4.9\times 10^3\ N/m$.

Explanation:

Mass of the rock is 2.5 kg

It is dropped off a 32 m cliff and hits a spring, compressing it 57 cm. It is required to find the spring constant.

When it falls form a height, it possess potential energy and when it hits a spring, the potential energy gets converted to elastic potential energy such that :

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

Here, k is spring constant

$k=\dfrac{2mgh}{x^2}\\\\k=\frac{2\cdot2.5\cdot10\cdot32}{0.57^{2}}\\\\k=4924.59\ N/m$

$k=4.9\times 10^3\ N/m$

So, the spring constant of the spring is $k=4.9\times 10^3\ N/m$.