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.

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