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

jupiter

explanation:

it is so massive and has such a strong gravitational pull its likely prevented the formation of a planet between itself and mars in the asteroid belt

subscribe to my yt channel yoongi jmz

follow my insta yoongi_jams

stan bts

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2