Question

A box is allowed to slide a distance d down a ramp a constant speed while being pushed by a horizontal force, Fa. The mass of the box is m. The ramp is inclined at an angle of above the horizontal. If the coecient of kinetic friction between the box and the ramp is k, how much work is done by the horizontal force as the box moves a distance d down the ramp? Your answer should only contain the parameters m, d, , k, and the physical constant g.

Answer 1

Let the incline angle of Ramp =  θ
As the Force is Horiozntal,
Resolution of force in direction along the incline and Perendicular to incline.

Force parallel to incline = Fa*cos(θ)
Force Perendicular to incline = Fa*sin(θ)

As the movement is only along the incline.
Work done = Force * displacement
Work done = Fa*cos(θ) * d

Calculation of Force Fa-

Friction Force = k*N
Normal Force , N = mg*cos(θ) + Fa*sin(θ)

Friction Force = k*( mg*cos(θ) + Fa*sin(θ))

Weight of box acting down parallel to the incline = mg*sin(θ)

Equation can be written as -
Fa*cos(θ) +  mg*sin(θ) - k*(  mg*cos(θ) + Fa*sin(θ)) = mass * acceleration

As box is moving with constant speed , acceleration = 0

Fa*cos(θ) +  mg*sin(θ) - k*(  mg*cos(θ) + Fa*sin(θ)) = 0
Fa*cos(θ) = k*(  mg*cos(θ) + Fa*sin(θ)) - mg*sin(θ))

Now Work done by Horizontal force =
W = k*(  mg*cos(θ) + Fa*sin(θ)) - mg*sin(θ)) * d