A crane lifts a 487 kg beam vertically at a constant velocity. If the crane does 5.20 x 10^4 of work on the beam find the vertical distance that it lifted the beam?

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I understand the equation that needs to be used to solve the problem. What don't I understand is how does the coefficient of gravity(g) which is 9.8m/s plays a role in this question. Can someone kindly explain this part for me?

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Answer 1 · 2016-05-08T13:03:31

Vertical distance $d = 10.9$ $d=10.9" "$ meters

The only force involve in the crane is the weight $F$ $F$ of the beam

The force of the weight
$F = m g$ $F=mg$

$F = 487$ $F=487" "$ $k g \cdot (- 9.8 \frac{m}{{sec}^{2}})$ $kg*(-9.8" "m/sec^2")$

$F = - 4772.6$ $F=-4772.6" "$ Newtons (negative because weight is downward direction)

Force of the crane is equal to weight directed upward (positive) $= + 4772.6$ $=+4772.6" "$ Newtons

The work $W = F \cdot d$ $W=F*d" "$ where $d =$ $d=$ distance

$d = \frac{W}{F} = \frac{5.2 x 10^{4}}{4772.6}$ $d=W/F=(5.2 x 10^4)/4772.6$

$d = 10.89552864$ $d=10.89552864" "$ meters

God bless....I hope the explanation is useful.