Answer:
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
Due to the friction force, the object will stop after traveling some distance. The distance traveled by the block before it comes to rest will be 13.98 m.
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
The given data in the problem is;
m is the mass of object =3kg
W is the Weight =mg =3×9.81=29.43N
Θ is the Inclination angle =30°
u is the Initial velocity below the ramp =16 m/sec
μk is the Coefficient of kinetic friction=0.5
μs is the Coefficient of static friction =0.8
The components of the forces are done on the inclined plane. On balancing the force in the y-direction.
[tex]\rm N = WCos\theta\\\\ \rm N=29.43Cos30^0\\\\ \rm N = 25.49N[/tex]
From Newton's second law of motion;
ΣF = ma
On balancing the horizontal components we got;
[tex]\rm F_x= WSin\theta \\\\ \rm F_x=29.43Sin30^0\\\\ F_x=14.715[/tex]
Cofficient of the kinetic friction is given as;
[tex]\rm F_K= \mu_k N \\\\ \rm 0.5 \times 25.49 \\\\ \rm F_K=12.745 N[/tex]
On applying Newton's second law of motion;
[tex]\rm \sum F=ma \\\\ -F_X-F_r= ma \\\\ \rm -14.715-12.745=3a \\\\ 3a= -27.46 \m/sec^2 \\\\ a= -9.15 \ m/sec^2[/tex]
If the acceleration is constant then from the second equation of motion;
[tex]\rm v^2 = u^2+2as \\\\ 0^2= 16^2+2\times 9.15s \\\\ 18.31 s=256 \\\\ s= 13.98\ m[/tex]
Hence the distance traveled by the block before it comes to rest will be 13.98 m.
To learn more about the friction force refer to;
https://brainly.com/question/1714663
