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Scorpion4ik [409]
4 years ago
10

Pete slides a crate up a ramp with constant speed at an angle of 24.3 ◦ by exerting a 289 N force parallel to the ramp. How much

work has been done against gravity when the crate is raised a vertical distance of 2.39 m? The coefficient of friction is 0.33.
Physics
1 answer:
Bumek [7]4 years ago
5 0

Answer:

W=972.73 J

Explanation:

Given that

θ = 24.3°

F= 289 N

h= 2.39 m

μ = 0.33

Given that velocity is constant is means that acceleration is zero that is why force will be balance.

Lest take mass = m

Gravity force along ramp = m g sinθ

Friction force =  μm g cosθ

F= m g sinθ  + μm g cosθ

By putting the values

F= m g (sinθ  + μcosθ)              

289 = 10 m  ( sin 24.3°+ 0.33 cos 24.3°)                 ( take g =10 m/s²)

289 = 10 m(0.71)

m = 40.7 kg

The force against gravity W

W= m g h

W= 40.7 x 10 x 2.39 J

W=972.73 J

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