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An object of mass m is lifted straight up at constant speed through a height h, so the lifting force equals its weight mg. What is the resulting change in gravitational potential energy of the object-Earth system?

ADelta PEg equals m plus g plus h
BDelta PEg equals m times g, divided by h
CDelta PEg equals h divided by m times g
DDelta PEg equals m times g times h
Answer & Solution
Correct answer: D. Delta PEg equals m times g times h
1. Lifting the object at constant speed requires a force equal to its weight, mg, and the work done on it is W = (force)(distance) = mgh. 2. This work done against gravity is defined to be the change in gravitational potential energy, so delta PEg = mgh. 3. Adding the three quantities together, or forming a ratio between them, does not match a definition built from force times distance. 4. Only the product of mass, g, and height correctly reproduces the work done in lifting the object. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.3 Gravitational Potential Energy_
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