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23.03.2020 •
Physics
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.)
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Ответ:
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
Ответ:
e
Explanation: