A 3.00-kg block is sent up a ramp of angle θ θ equal to 30.0° with an initial velocity ν0 ν 0 equal to 16.0 m/s. Between the block and the ramp, the coeffiient of kinetic friction is μk μ k equal to 0.50 and the coeffiient of static friction is μs μ s equal to 0.80. 1) How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop? (Express your answer to two significant figures.)

Explanation:

Given that,

Mass of object =3kg

Weight W=mg =3×9.81=29.43N

Inclination angle =30°

Initial velocity below the ramp is

Vi=16m/s

Coefficient of kinetic friction μk=0.5

Coefficient of static friction μs=0.8

To know the acceleration of the body up the ramp

For the vertical direction, the two reactions are the normal and the weight resolved to vertical direction

Then, N = WCosθ

N=29.43Cos30

N = 25.49N

For horizontal direction

Let use newton law of motion

ΣF = ma

Forces acting on the body going up the ramp is

1. The horizontal component of the weight, which is acting downward and it value is WSinθ.

Fx= WSinθ

Fx=29.43Sin30

Fx=14.715N

2. The frictional force which opposes motion and it is directed downward, it value is

Fr=μk•N

Fr=0.5×25.49

Fr=12.745N

Now applying the equation

ΣF = ma

-Fx-Fr=ma

-14.715-12.745=3a

-27.46=3a

a=-27.46/3

a=-9.15m/s²

This show that body is decelerating

Now using equation of motion to find the distance travelled

Final velocity is zero Vf=0

Vf²=Vi²+2as

0²=16²+2•-9.15•s

0=256-18.31s

18.31s=256

s=256/18.31

s=13.98m

So the distance traveled is 13.98m

Check attachment for further analysis and diagram

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