Work Done by the Gravitational Force — Question 2

Question 2

A lifting agent must supply both gravitational and kinetic requirements

Take a uniform gravitational field with $g=10$ m/s$^2$, upward y positive, and neglect air resistance. A 5 kg load is lifted vertically by 4 m, starting from rest and ending at 3 m/s upward. Only gravity and a lifting force do work.

1. Find gravitational work and the kinetic-energy change.

2. Find work by the lifting force.

3. Find the displacement-averaged lifting force and explain why it need not be the force at every instant.

Question 2 - Solution

A lifting agent must supply both gravitational and kinetic requirements

(a) Separate the two known changes.

The governing relation is $$\boxed{W_{\mathrm{lift}}+W_g=\tfrac12m(v_f^2-v_i^2)}$$

$$W_g=-5(10)(4)=-200\ \mathrm J.$$

$$\Delta K=\tfrac12(5)(3^2)=22.5\ \mathrm J.$$

(b) Use total work.

$$W_{\mathrm{lift}}+W_g=\Delta K,$$ so $$W_{\mathrm{lift}}=222.5\ \mathrm J.$$

(c) Divide the work by the upward distance.

The force component averaged over displacement is $$222.5/4=55.625\ \mathrm N.$$ Endpoint data fix its integral, not its time history. This equals the constant lifting force only if constancy is additionally specified.