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You push on a 30.0-kg package with a constant applied force of 120 N through a distance of 0.800 m, while an opposing friction force averages 5.00 N. What is the net work done on the package?

A96.0 J
B100 J
C92.0 J
D4.00 J
Answer & Solution
Correct answer: C. 92.0 J
1. The net force is the applied force minus the opposing friction force, Fnet = 120 N minus 5.00 N, which is 115 N. 2. Net work is Wnet = Fnet times d. 3. Substituting the values, Wnet = (115 N)(0.800 m). 4. That multiplication gives Wnet = 92.0 J. 5. 96.0 J is the work from the applied force alone, ignoring friction, and 4.00 J is just the magnitude of the friction force's own work, not the combined net work; 100 J does not follow from either force correctly. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.2 Kinetic Energy and the Work-Energy Theorem_
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