A man exerts a constant force to pull a 51-kg box across a floor at constant speed. He exerts this force by attaching a rope to
the box and pulling so that the rope makes a constant angle of 36.9∘ above the horizontal. The coefficient of kinetic friction for the box-floor interface is μk = 0.12. What is the work done by the man if he moves the box 10m
1 answer:
Answer:
W=561.41 J
Explanation:
Given that
m = 51 kg
μk = 0.12
θ = 36.9∘
Lets F is the force applied by man
Given that block is moving at constant speed it mans that acceleration is zero.
Horizontal force = F cos θ
Vertical force = F sinθ
Friction force Fr= μk N
N + F sinθ = m g
N = m g - F sinθ
Fr = μk (m g - F sinθ)
For equilibrium
F cos θ = μk (m g - F sinθ)
F ( cos θ +μk sinθ) = μk (m g
Now by putting the values
F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10
F= 70.2 N
We know that Work
W= F cos θ .d
W= 70.2 x cos 36.9∘ x 10
W=561.41 J
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