Question

A vertical ideal spring of constant 42 N/m is placed on top of a lab bench. A block of unknown mass, m, is dropped onto the spring from a height of 1.6 m above the lab bench. Before the block hits the spring, the spring is 0.3 m long. Once the block comes to rest after hitting the spring, the spring is now 0.09 m long. What is the mass of the block?

Answer 1

The potential energy of the block before it hits the spring is given by:

PE = mgh

where m is the mass of the block, g is the acceleration due to gravity, and h is the height from which the block is dropped.

PE = mgh = m × 9.8 m/s² × 1.6 m = 15.68m J

When the block comes to rest after hitting the spring, the potential energy is converted into the elastic potential energy stored in the spring. The elastic potential energy of the spring is given by:

EPE = (1/2)kx²

where k is the spring constant and x is the extension of the spring from its unstretched length.

The change in potential energy of the block is equal to the elastic potential energy stored in the spring, so:

15.68m J = (1/2)(42 N/m)(0.09 m - 0.3 m)²

15.68m J = 2.8356 J

m = 2.8356 J / 15.68 J/kg = 0.1807 kg

Therefore, the mass of the block is approximately 0.1807 kg.