Conservation of Mechanical Energy — Question 5

Question 5

A counterweight pair opposed by a spring

Neglect friction and air resistance, take $g=10$ m/s$^2$, and treat all stated guides as fixed. Masses 2 kg and 3 kg hang from an ideal cord over a massless frictionless fixed pulley. The 3 kg mass is also attached below to a vertical spring of $k=40$ N/m, initially relaxed. It descends by x and compresses the spring by x. Both masses start at rest.

1. Find total kinetic energy as a function of x.

2. Find the next stopping displacement.

3. Find the position and value of maximum common speed.

Question 5 - Solution

A counterweight pair opposed by a spring

(a) Sum both gravitational changes and the spring store.

Both masses contribute to the gravitational change: $$\Delta U_g=2gx-3gx=-10x,\qquad\Delta U_s=\tfrac12(40)x^2=20x^2.$$ $$\boxed{K_{\mathrm{total}}=-(\Delta U_g+\Delta U_s)=10x-20x^2}.$$

(b) Find the positive zero.

The next stop is the nonzero root: $$K=x(10-20x)=0\Rightarrow\boxed{x=.50\ \mathrm m}.$$ The inextensible cord makes both masses stop together.

(c) Maximize total K and include both inertias.

Maximize the kinetic energy: $$\frac{dK}{dx}=10-40x=0\Rightarrow\boxed{x=.25\ \mathrm m},\qquad K_{\max}=1.25\ \mathrm J.$$ $$\tfrac12(2+3)v_{\max}^2=1.25\Rightarrow\boxed{v_{\max}=\sqrt{.50}=.7071\ \mathrm{m/s}}.$$ Both moving masses contribute to the inertia.