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.

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Quizplus · Accepted Answer

The work done by the crane on the beam can be calculated using the formula $W = Fd$, where $F$ is the force applied and $d$ is the distance moved in the direction of the force. Here, the force $F$ includes not only the force to counteract gravity (the weight of the beam, $mg$), but also the force to provide the upward acceleration ($ma$). Thus, $F = mg + ma$, where $m = 425 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $a = 1.8 \, \text{m/s}^2$. The distance $d = 95 \, \text{m}$.Substituting the values, we get:$$F = 425 \times 9.8 + 425 \times 1.8 = 425 \times (9.8 + 1.8) = 425 \times 11.6 = 4930 \, \text{N}$$Then, calculating the work done:$$W = Fd = 4930 \times 95 = 468350 \, \text{J} = 4.7 \times 10^5 \, \text{J}$$

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 $95 \mathrm {~m} .$