Question

A Block on a frictionless floor is attached to the free end of a spring. An applied force of magnitude F is equal to 4.9N and would be needed to hold the block stationary at X_Initial = 0.012metres. If the block is moved leftward to X_Final = -0.012metres, how much work does the spring force do on the block during this displacement? Explain the sign of this work.

Answer 1

$$F=kx_i\\4.9=0.012k\\k=408\frac{N}{m}$$

$$-W=0.5kx_f^2+0.5kx_i^2\\W=-408(0.012)^2=-0.059\ J$$

Negative sign because we apply force in the negative x-direction.