A 20 -kg block linitially sliding at constant speed on a frictionless floor starts cornpressing a spring from the spring's equilibrium position. How much work does the spring force do on the block as the spring is compressed by \( d=30 \mathrm{~cm} \), as shown? Assurme that the spring is ideal and massless and that its force constant is \( 200 \mathrm{~N} / \mathrm{m} \).

To calculate the work done by the spring force as it compresses by \( d = 30 \, \text{cm} \) (or \( 0.3 \, \text{m} \)), we can use the formula for the work done by a spring, which is given by: \[ W = -\frac{1}{2} k d^2 \] where \( k \) is the spring constant and \( d \) is the compression distance. Plugging in your values: \[ W = -\frac{1}{2} (200 \, \text{N/m}) (0.3 \, \text{m})^2 = -\frac{1}{2} (200) (0.09) = -9 \, \text{J} \] So, the spring does \( 9 \, \text{J} \) of work on the block, and the negative sign indicates that the work is done opposite to the direction of compression. Now, when the spring compresses, it stores potential energy, which means that as it pushes back, it transforms that energy into kinetic energy of the block (unless there’s friction). So, if your block stops at maximum compression, you’ll see that the work done on it causes it to slow down. It's like trying to shove a soft toy into a small box! The more you push, the more it resists, and some energy gets stored in the spring action—just waiting to bounce back!