What is the force exerted by the spring on the block?

What are the given parameters for calculating the force exerted by the spring on the block?

The given information states that the block has a mass of 60 grams, the spring stiffness is 13 N/m, the relaxed length of the spring is 0.6 m, and the spring is stretched to 0.75 m before being released. We are also told that the friction is negligible and that we can approximate the force on the block by the spring as constant.

Calculating the Force Exerted by the Spring

Given: The mass of the block is 60 grams, the spring stiffness is 13 N/m, the relaxed length of the spring is 0.6 m, and the spring is stretched to 0.75 m before being released.

To find the force exerted by the spring on the block, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. The equation for Hooke's Law is F = kx, where F is the force, k is the spring stiffness, and x is the displacement.

In this case, the displacement of the spring is 0.75 m - 0.6 m = 0.15 m. Substituting this value into the equation, we get F = (13 N/m)(0.15 m) = 1.95 N.

Therefore, the force exerted by the spring on the block is approximately 1.95 N.