Abox 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)

 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

      = U = mg y

We use trigonometry to find the height (y)

     sin 21.8 = y / d

     y = d sin 21.8

      = m g d sin 21.8

Let's replace

     W = ΔEm =   - 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

its a 5.5 n

glad that

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .