A crane lifts a 425 kg steel beam vertically a distance of 117 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

Work done by the crane is equal to the change in kinetic energy of the beam.$$	ext{Work done}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$$$=\frac{1}{2}(425	ext{ kg})(1.8	ext{ m/s})^2-0$$$$=5.8	imes10^5	ext{ J}$$

A Crane Lifts a 425 Kg Steel Beam Vertically a Distance