Question

A 20 kg box on a horizontal frictionless surface is moving to the right at a speed of 4.0 m/s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.50 m, and the box then oscillates back and forth. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero. ____ Positive ____ Negative ____ Zero

Answer 1

Answer:

Work done by the spring is negative

Explanation:

We can answer this question by thinking what is the force acting on the box.

In fact, the force acting on the box is the restoring force of the spring, which is given by Hooke's Law:

$F=-kx$

where

k is the spring constant

x is the displacement of the box with respect to the equilibrium position of the spring

The negative sign in the equation indicates that the direction of the force is always opposite to the direction of the displacement: so, whether the spring is compressed or stretched, the force applied by the spring on the box is towards the equilibrium position.

The work done by the restoring force is also given by

$W=Fx cos \theta$

where

F is the restoring force

x is the displacement

$\theta$ is the angle between the direction of the force and the displacement

Here we know that the force is always opposite to the displacement, so

$\theta=180^{\circ} \rightarrow cos \theta =-1$

Which means that the work done by the spring is always negative, since the direction of the restoring force is always opposite to the direction of motion.