How much work is needed to push a 117- kg packing crate a distance of 2.75 m up an inclined plane that makes an angle of 28 o with the horizontal, when inclined plane is frictionless

Explanation:

The minimum total work is the work needed to counteract the work associated with the weight:

Given Information:  

mass of crate = m = 117 kg

distance = d = 2.75 m  

Inclined plane angle = θ = 28°

Required Information:  

Work done = W = ?

Work done = 1480.3 N.m

Explanation:

Refer to the attached image, since there is no friction, the only force acting on the crate is the force in the downward direction given by

Fy = mgsin(θ)

Fy = 117*9.8*sin(28°)

Fy = 538.3 N

Now as we know work done is given by

W = Fy*d

W = 538.3*2.75

W = 538.3*2.75

W = 1480.3 N.m

Therefore, 1480.3 N.m of work is required to push a 117 kg  of crate to a distance of 2.75 m up an inclined plane making an angle of 28° with the horizontal surface.

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