Question

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

Answer 1

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