Work and Kinetic Energy — Question 5

Question 5

Endpoint speeds can reveal an unmeasured resistance

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. A 2 kg block moves 5 m horizontally right. Its speed rises from 2 to 6 m/s while a constant 20 N pull acts $30^\circ$ above horizontal. A constant horizontal resistance opposes the motion. No other force does work.

1. Find the net work required by the speed change.

2. Find the work and magnitude of the resistance.

3. Check whether a proposed resistance of 12 N is compatible with the measured speeds.

Question 5 - Solution

Endpoint speeds can reveal an unmeasured resistance

(a) Convert both endpoint speeds to energies.

The governing relation is $$\boxed{\tfrac12m(v_f^2-v_i^2)=Fd\cos\theta-Rd}$$

$$W_{\mathrm{net}}=\tfrac12(2)(36-4)=32\ \mathrm J.$$

(b) Subtract the known pull contribution.

$$W_P=20(5)\cos30^\circ=50\sqrt3\ \mathrm J.$$ Thus $$W_R=32-50\sqrt3\approx-54.603\ \mathrm J,\qquad R=(50\sqrt3-32)/5=10\sqrt3-6.4\approx10.921\ \mathrm N.$$

(c) Predict rather than refit.

If R=12 N, net work is $50\sqrt3-60\approx26.603\ \mathrm J$. Then $$K_f=4+26.603=30.603\ \mathrm J$$ and speed is about 5.532 m/s, not 6 m/s. The proposal conflicts with the observations.