Question

You push on a spring whose stiffness is 100 N/m, compressing it until it is 3 cm shorter than its relaxed length. The magnitude of the force the spring now exerts on your hand is: ∣F₈∣=?

Answer 1

When a spring with a stiffness of 100 N/m is compressed by 3 cm, it exerts a force of 3 Newtons. The negative sign in Hooke's Law equation means the force is in the opposite direction of the displacement.

Step-by-step explanation:

The question asks about the force exerted by a spring when compressed. According to Hooke's Law, the force (F) exerted by a spring is proportional to the displacement from its equilibrium position, represented by the equation F = -kx, where k is the spring constant and x is the displacement. In this case, the stiffness of the spring is 100 N/m, and it's compressed 3 cm, which is 0.03 m. To find the magnitude of the force, we ignore the negative sign and use the formula:

|F| = kx = (100 N/m)(0.03 m) = 3 N.

Therefore, the spring exerts a force of 3 Newtons on your hand. The negative sign in the Hooke's Law equation indicates that the restoring force is in the direction opposite the displacement, which means it's a restoring force that tries to bring the spring back to its equilibrium position.