Question

A crane lifts a 425 kg steel beam vertically a distance of 66 m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s2? Neglect frictional forces

Answer 1

\[W=F*d\]
\[F=ma\]
\[W=(ma)*d\]
\[W=((425 kg)(\frac{1.8 m}{s^2}))*66m\]