Work and Kinetic Energy — Question 1

Question 1

The theorem requires net work rather than the largest contribution

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. A 2 kg cart initially moves right at 3 m/s. Over a 5 m rightward displacement a horizontal drive does +60 J of work and resistance does $-20$ J. Weight and normal force do zero work; there are no other forces.

1. Find the net work and final kinetic energy.

2. Find the final speed.

3. What false speed follows if resistance is omitted, and why is it too large?

Question 1 - Solution

The theorem requires net work rather than the largest contribution

(a) Add every contribution.

The governing relation is $$\boxed{K_f=K_i+\sum_jW_j,\qquad v_f=\sqrt{v_i^2+2W_{\mathrm{net}}/m}}$$

$$W_{\mathrm{net}}=60-20=40\ \mathrm J.$$ Initial energy is $$K_i=\tfrac12(2)3^2=9\ \mathrm J,$$ so $$K_f=49\ \mathrm J.$$

(b) Recover the nonnegative speed.

$$v_f=\sqrt{2K_f/m}=7\ \mathrm{m/s}.$$

(c) Audit the incomplete force inventory.

Omitting resistance gives $$K_f=69\ \mathrm J,\qquad v_f=\sqrt{69}\approx8.307\ \mathrm{m/s}.$$ The missing negative work removes energy, so ignoring it exaggerates the speed.