# 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?

## 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?

Vertical distance $d = 10.9 \text{ }$meters

#### Explanation:

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

The force of the weight
$F = m g$

$F = 487 \text{ }$$k g \cdot \left(- 9.8 \text{ "m/sec^2}\right)$

$F = - 4772.6 \text{ }$Newtons (negative because weight is downward direction)

Force of the crane is equal to weight directed upward (positive)$= + 4772.6 \text{ }$Newtons

The work $W = F \cdot d \text{ }$where $d =$distance

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

$d = 10.89552864 \text{ }$meters

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