A crane lifts a 425 kg steel beam vertically upward a distance of 95 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 can be calculated using the formula $W = Fd$, where $F$ is the force applied and $d$ is the distance moved. The force $F$ can be found by using $F = ma + mg$, where $m$ is the mass of the beam, $a$ is the acceleration, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Substituting the given values, $F = 425 \times (1.8 + 9.8) = 425 \times 11.6 = 4930 \, \text{N}$. The work done, $W$, is then $4930 \times 95 = 468350 \, \text{J}$, which is approximately $4.7 \times 10^5 \, \text{J}$.

A Crane Lifts a 425 Kg Steel Beam Vertically Upward 3.2×105 J3.2 \times 10 ^ { 5 } \mathrm {~J}3.2×105 J

A Crane Lifts a 425 Kg Steel Beam Vertically Upward $3.2 \times 10 ^ { 5 } \mathrm {~J}$