Question

A block of mass m = 3 kg is released from rest on a frictionless incline of angle �� = 25 ��. The block collides with a massless platform that is attached to the free end of a spring on the incline as shown. The maximum compression of the spring after the collision is d = 0.23 m. The spring constant is k = 800 N/m. The height of the incline is h = 0.11 m. What is the potential energy of the block-spring system when the spring is compressed by 0.23 m?

Answer 1

The potential energy of the block-spring system when the spring is compressed by 0.23 m is 20.92 J.

Step-by-step explanation:

To find the potential energy of the block-spring system when the spring is compressed by 0.23 m, you can use the formula for potential energy of a spring:

Potential Energy = (1/2) * k * x^2

where k is the spring constant and x is the compression or extension of the spring.

Plugging in the values, the potential energy of the block-spring system is:

Potential Energy = (1/2) * 800 N/m * (0.23 m)^2 = 20.92 J