Work and Kinetic Energy — Question 4

Question 4

Zero net work permits a changing velocity vector

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. A 2 kg particle is guided at constant speed 5 m/s through a quarter circle in a horizontal plane. It starts with velocity $(5,0)$ m/s and ends with $(0,5)$ m/s. The guide force is everywhere perpendicular to the motion.

1. Find the kinetic-energy change and net work.

2. Find the change in velocity.

3. Why does zero net work not imply zero net force during the turn?

Question 4 - Solution

Zero net work permits a changing velocity vector

(a) Compare the equal speed magnitudes.

The governing relation is $$\boxed{W_{\mathrm{net}}=\Delta K=\tfrac12m(v_f^2-v_i^2)}$$

$$K_i=K_f=\tfrac12(2)25=25\ \mathrm J,$$ so $$\Delta K=W_{\mathrm{net}}=0.$$

(b) Retain directional information.

$$\Delta\vec v=(-5,5)\ \mathrm{m/s},\qquad |\Delta\vec v|=5\sqrt2\ \mathrm{m/s}.$$

(c) Distinguish tangential from normal acceleration.

The nonzero inward force changes velocity direction while remaining perpendicular to each infinitesimal displacement. It does no work. The theorem constrains changes in speed, not every component of acceleration.