Question

A crane lifts a 425 kg steel beam vertically upward 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

Answer:

W = 311074.5 [J]

Explanation:

In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.

∑F = m*a

F = forces acting on the beam [N]

m = mass = 425 [kg]

a = acceleration = 1.8 [m/s²]

The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)

$F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]$

Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.

$W=F*d$

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

$W=4713.25*66\\W=311074.5[J]$