Question

A 15 kg runaway grocery cart runs into a spring with spring constant 260 N/m and compresses it by 58 cm. What is the force exerted by the spring on the cart?

1) 150 N
2) 260 N
3) 3770 N
4) 7540 N

Answer 1

Using Hooke's Law, the force exerted by the spring on the runaway grocery cart is calculated to be 150.8 N, which rounds to 150 N, making option 1) the correct answer.

Step-by-step explanation:

The question asks about the force exerted by the spring on a runaway grocery cart that has compressed the spring. To determine this force, we use Hooke's Law, which states that the restoring force exerted by a spring is directly proportional to the displacement from the equilibrium position, and is given by F = kx, where F is the force in newtons, k is the spring constant in newtons per meter, and x is the displacement in meters. In this case, the spring constant k is given as 260 N/m and the displacement x is 58 cm (which is 0.58 meters when converted). So we compute the force as F = 260 N/m × 0.58 m = 150.8 N, which rounds to 150 N, making option 1) the correct answer.