A 5.0-kg box is pushed up with initial velocity and travels 5.0 m up a plane that is inclined at 30.0^o with the horizontal before it stops. The coefficient of kinetic friction between the box and the plane is 0.20. The initial velocity of the block is: A) 5.8 m/s B) 8.1 m/s C) 4.0 m/s D) 66 m/s E) 16 m/s

Question

Quizplus · Accepted Answer

We can use the conservation of energy principle to solve the problem. The initial kinetic energy of the box is given by:

K1 = (1/2)mv^2

Where m is the mass of the box and v is its initial velocity. This energy is transformed into potential energy as the box moves up the incline, against the force of friction. The potential energy gained by the box is given by:

U = mgh

Where h is the height the box travels and g is the acceleration due to gravity. Therefore, we have:

K1 = U + Wfriction

Where Wfriction is the work done against friction. The work done is given by:

Wfriction = μkNcosθd

Where μk is the coefficient of kinetic friction, N is the normal force, θ is the angle of inclination, and d is the displacement along the incline. We can express N and cosθ in terms of the weight of the box:

N = mgcosθ
cosθ = sqrt(3)/2

Therefore,

Wfriction = (0.2)(5kg)(9.81m/s^2)(sqrt(3)/2)(5m)

Wfriction = 12.16 J

Now we can solve for the initial velocity of the block:

(1/2)(5kg)v^2 = (5kg)(9.81m/s^2)(5m) + 12.16 J

v^2 = 98.1m^2/s^2 + 24.32m^2/s^2

v^2 = 122.42m^2/s^2

v = 11.06m/s

The closest answer to this value is C) 4.0 m/s.

A 50-Kg Box Is Pushed Up with Initial Velocity and Travels