A crane lifts a 425 kg steel beam vertically upward a distance of $95 \mathrm {~m} .$ How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s²? Neglect frictional forces.

Question

Quizplus · Accepted Answer

The work done by the crane is equal to the change in kinetic energy of the beam. The initial kinetic energy of the beam is zero, and the final kinetic energy is:$$K_f = \frac{1}{2}mv^2 = \frac{1}{2}(425 	ext{ kg})(1.8 	ext{ m/s})^2 = 729 	ext{ J}$$The work done by the crane is:$$W = K_f - K_i = 729 	ext{ J} - 0 = 729 	ext{ J}$$Therefore, the correct answer is A.

A Crane Lifts a 425 Kg Steel Beam Vertically Upward 95 m.95 \mathrm {~m} .95 m.

A Crane Lifts a 425 Kg Steel Beam Vertically Upward $95 \mathrm {~m} .$