Answer:
[tex]W_{tot,min} = 1481.372\,J[/tex]
Explanation:
The minimum total work is the work needed to counteract the work associated with the weight:
[tex]W_{tot, min} = W_{F}[/tex]
[tex]W_{tot,min} = m\cdot g\cdot \sin \theta \cdot \Delta s[/tex]
[tex]W_{tot,min} = (117\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 28^{\textdegree}) \cdot (2.75\,m)[/tex]
[tex]W_{tot,min} = 1481.372\,J[/tex]
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.
