A small block of mass m = 0.130 kg is dropped from height h measured from the top of the spring of spring constant k = 370 N/m, which is originally in equilibrium. When the block hits the mass it compresses it by 2.5 cm. Determine the height from which the spring was dropped.
A) 0.12 m
B) 0.091 m
C) 0.066 m
D) 0.025 m
E) 0.16 m

Question

Quizplus · Accepted Answer

The potential energy of the block before it is dropped is converted into elastic potential energy of the spring after it hits the mass. Thus, we can equate the potential energies:
$mgh = \frac{1}{2}kx^2$
where x is the compression of the spring and h is the initial height from which the block is dropped.

Solving for h, we get:
$h = \frac{kx^2}{2mg} + x$

Substituting the given values, we get:
$h = \frac{(370 N/m)(0.025 m)^2}{2(0.130 kg)(9.81 m/s^2)} + 0.025 m$
$h = 0.066 m$

Therefore, the height from which the spring was dropped is 0.066 m, which corresponds to choice C.

A Small Block of Mass M = 0