Starting from W=Fs and weight=mg, derive ΔE_P=mgΔh. State the physical conditions used in the derivation. Explain the signs of gravitational energy change and gravity's work when the body is lowered.

Question

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Steady lifting gives F=mg and no kinetic-energy change. W=FΔh=mgΔh becomes the potential-energy increase when no energy is dissipated. Lowering gives Δh<0 and ΔE_P<0; gravity does positive work. Choose vertical displacement in a uniform field. Use steady speed to identify lifting force as mg. Substitute into W=Fs and identify s=Δh. State no dissipative loss/no kinetic-energy change in the derivation. Give opposite signs for potential-energy change and gravity's work during lowering.

Derive energy from a steady lift

Starting from W=Fs and weight=mg, derive ΔE_P=mgΔh. State the physical conditions used in the derivation. Explain the signs of gravitational energy change and gravity's work when the body is lowered.

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Your self-assessment

Starting from W=Fs and weight=mg, derive ΔEP=mgΔh. State the physical conditions used in the derivation. Explain the signs of gravitational energy change and gravity's work when the body is lowered.

What a complete response should do

Your self-assessment

Starting from W=Fs and weight=mg, derive ΔE_P=mgΔh. State the physical conditions used in the derivation. Explain the signs of gravitational energy change and gravity's work when the body is lowered.