Physics Question of the Day

1. $+240\,\text{J}$ (correct answer)
2. $+120\,\text{J}$
3. $-240\,\text{J}$
4. $+480\,\text{J}$

Explanation: This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). The sled has mass m = 5.0 kg, initial velocity v_i = 2.0 m/s, and final velocity v_f = 10.0 m/s. The net work done is W_net = ΔKE = ½m(v_f² - v_i²) = ½(5.0 kg)((10.0 m/s)² - (2.0 m/s)²) = ½(5.0)(100 - 4) = ½(5.0)(96) = 240 J. Since the sled speeds up (v_f > v_i), the net work is positive (+240 J), indicating that energy was added to the system. Choice A (+240 J) is correct because it properly applies the work-energy theorem with the correct ½ factor and velocity squared terms. Choice B (+120 J) is incorrect—it appears to result from forgetting to square the velocities, calculating something like ½m(v_f - v_i) × (v_f + v_i) = ½(5.0)(8)(12) = 240 J but then dividing by 2 again. When calculating net work from velocity changes: (1) calculate initial kinetic energy: KE_i = ½mv_i², (2) calculate final kinetic energy: KE_f = ½mv_f², (3) find net work as W_net = KE_f - KE_i, which is positive if the object speeds up and negative if it slows down.