Physics

Physics

Problem

Which expression gives the change in gravitational potential energy of an object of mass mm that is displaced vertically by Δy\Delta y near Earth’s surface, if upward is taken as the positive direction? a) ΔU=12mg Δy\Delta U = \tfrac12 m g\,\Delta y b) ΔU=−mg Δy\Delta U = - m g\,\Delta y c) ΔU=mg Δy\Delta U = m g\,\Delta y d) ΔU=mg Δy+12mv2\Delta U = m g\,\Delta y + \tfrac12 m v^{2}
Intuitive explanation

Gravitational potential energy depends on height, and the key is tracking the sign carefully. When an object moves upward (positive Δy\Delta y), it moves against gravity's pull, so it gains potential energy—the change should be positive. The expression ΔU=mgΔy\Delta U = mg\Delta y captures this: gravity's strength (mgmg) times the vertical displacement, with the sign naturally following your choice of positive direction.

Intuitive explanation

Gravitational potential energy depends on height, and the key is tracking the sign carefully. When an object moves upward (positive Δy\Delta y), it moves against gravity's pull, so it gains potential energy—the change should be positive. The expression ΔU=mgΔy\Delta U = mg\Delta y captures this: gravity's strength (mgmg) times the vertical displacement, with the sign naturally following your choice of positive direction.

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