Question

A spring is stretched 0.200 m. If the spring constant is 200 N/m, what is the weight of the block?

1) 20 N
2) 8 N
3) 4 N
4) 40 N

Answer 1

The weight of the block attached to a spring with a spring constant of 200 N/m that is stretched 0.200 m is 40 N.

Step-by-step explanation:

The weight of the block when a spring with a spring constant of 200 N/m is stretched by 0.200 m can be found using Hooke's Law, which states that the force required to stretch or compress a spring by a distance x from its natural length is proportional to x. The formula is F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. In this case, k = 200 N/m and x = 0.200 m, therefore F = 200 N/m × 0.200 m = 40 N. Since the force exerted by the spring is equivalent to the weight of the block, the weight of the block is 40 N.