Conservation of Mechanical Energy — Question 1

Question 1

Speed on a track with intervening valleys

Neglect friction and air resistance, take $g=10$ m/s$^2$, and treat all stated guides as fixed. A 2 kg bead is constrained to a smooth fixed wire. At A, height 5 m, its speed is 2 m/s. It later reaches B at height 2 m after passing through a valley. The wire has no point above A between them.

1. Find mechanical energy with U=0 at height zero.

2. Find speed at B.

3. Explain why valley depth and normal-force magnitude do not enter this speed calculation.

Question 1 - Solution

Speed on a track with intervening valleys

(a) Include the initial kinetic energy.

Include both initial energy stores: $$\boxed{E=K_A+U_A=\tfrac12(2)(2)^2+2(10)(5)=104\ \mathrm J}.$$

(b) Subtract the final potential.

At B, $U_B=2(10)(2)=40$ J, hence $$K_B=104-40=64\ \mathrm J,\qquad\boxed{v_B=\sqrt{2K_B/m}=8\ \mathrm{m/s}}.$$

(c) Use the work of the constraint.

The fixed smooth constraint does no work: $$W_N=\int\vec N\cdot d\vec r=0,\qquad E=\text{constant}.$$ Valley depth changes intermediate speeds and forces, not the final energy difference. Since $U\leq100$ J along the route, $K\geq4$ J and no energy barrier intervenes.