A 20 kg Box on a Horizontal Frictionless Surface Attached to a Spring

Answer

(a)(i) Sign of the spring's work

The work is negative. During compression the box moves to the right while the spring force points to the left, opposite the displacement. The box also slows from $4.0\,\mathrm{m/s}$ to zero, so its kinetic-energy change is negative.

(a)(ii) Magnitude of the work

By the work-energy theorem,

$$W_{\mathrm{spring}}=\Delta K =0-\frac12mv^2 =-\frac12(20)(4.0)^2 =-160\,\mathrm{J}.$$

The requested magnitude is $\boxed{|W_{\mathrm{spring}}|=160\,\mathrm{J}}$.

(b) Spring constant

With no friction, the initial kinetic energy becomes spring potential energy at maximum compression:

$$\frac12mv^2=\frac12kx_{\max}^2.$$

Thus

$$k=\frac{mv^2}{x_{\max}^2} =\frac{(20)(4.0)^2}{(0.50)^2} =\boxed{1.28\times10^3\,\mathrm{N/m}}.$$

(c) Maximum acceleration

The spring force has its largest magnitude at maximum displacement:

$$a_{\max}=\frac{kx_{\max}}{m} =\frac{(1280)(0.50)}{20} =\boxed{32\,\mathrm{m/s^2}}.$$

(d) Oscillation frequency

For an ideal mass-spring oscillator,

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}} =\frac{1}{2\pi}\sqrt{\frac{1280}{20}} =\frac{8}{2\pi} \approx\boxed{1.27\,\mathrm{Hz}}.$$