Question

A 10.0 kg mass is placed on a frictionless, horizontal surface. The mass is connected to the end of a horizontal compressed spring which has a spring constant 339 N/m. When the spring is released, the mass has an initial, positive acceleration of 10.2 m/s2. What was the displacement of the spring, as measured from equilibrium, before the block was released? Watch the sign of your answer.

Answer 1

Answer:

The displacement of the spring, as measured from equilibrium, is 0.301 m

Explanation:

Hi there!

Using Hooke´s law, we can calculate the displacement of the spring:

F = -kx

Where:

F = restoring force exerted by the spring.

k = spring constant.

x = displacement of the spring.

The force exerted by the spring can also be calculated using the Newton law:

F = m · a

Where:

F = force.

m = mass of the object.

a = acceleration.

Then, combining both laws:

F = -k · x = m · a

-k · x = m · a

-339 N/m · x = 10.0 kg · 10.2 m/s²

x = 10.0 kg · 10.2 m/s² / -339 N/m

x = -0.301 m

The displacement of the spring, as measured from equilibrium, is 0.301 m