Question

A box of mass m = 23.0 kg is sliding along a horizontal frictionless surface at a speed vi = 7.10 m/s when it encounters a ramp inclined at an angle of θ = 21.8°. The coefficient of kinetic friction between the ramp and the box is μ = 0.0704 and the box slides a distance d up the ramp before coming momentarily to rest.
Determine the distance the box slides up the ramp before coming momentarily to rest. (in meters)

Answer 1

Answer:

 d = 57.7 m

Explanation:

For this exercise we will use energy work theorems

     W = ΔEm

Let's calculate the work of the rubbing force on the ramp

    W = -fr d

The negative sign is because the force of friction opposes the movement

Newton's second law in the axis perpendicular to the ramp gives

    N- W cos 21.8 = 0

    N = mg cos 21.8

The equation for the force of friction is

    fr = μN

    fr = μ mg cos 21.8

Let's replace

    W = - μ mg d cos 21.8

Let's look for energy in two points

Initial just before entering the ramp

     Emo = K = ½ m v²

End the highest point where the body stops

     $Em_{f}$ = U = mg y

We use trigonometry to find the height (y)

     sin 21.8 = y / d

     y = d sin 21.8

      $Em_{f}$ = m g d sin 21.8

Let's replace

     W = ΔEm =  $Em_{f}$ - Em₀

    - μ mg d cos 21.8 = mgd sin21.8 - ½ m v²

    m g d (sin21.8 + μ cos 21.8) = ½ m v²

    d = v² / (2 (sin21.8 + μ cos21.8))

    d = 7.10² / (2 (sin21.8 + 0.0704 cos 21.8))

    d = 50.41 /0.87345

    d = 57.7 m