Question

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

Answer 1

Given Information:  

mass of crate = m = 117 kg

distance = d = 2.75 m  

Inclined plane angle = θ = 28°

Required Information:  

Work done = W = ?

Answer:  

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.

Answer 2

Answer:

$W_{tot,min} = 1481.372\,J$

Explanation:

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

$W_{tot, min} = W_{F}$

$W_{tot,min} = m\cdot g\cdot \sin \theta \cdot \Delta s$

$W_{tot,min} = (117\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 28^{\textdegree}) \cdot (2.75\,m)$

$W_{tot,min} = 1481.372\,J$