A 15.0 kg stone slides down a snow-covered hill, leaving point A at a speed of 10.0 m/s
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Question “A 15.0 kg stone slides down a snow-covered hill, leaving point A at a speed of 10.0 m/s”
A 15.0 kg stone slides down a snow-covered hill, leaving point A at a speed of 10.0 m/s. There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.0 N/m. The coefficients of kinetic friction between the stone and the horizontal ground is 0.2.
A. What is the speed of the stone when it reaches point B?
B. How far will the stone compress the spring?
Answer
Answer 1
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Answer 2
Given:
Stone mass, m=15 kg
Initial velocity of stone, i.e. Velocity of the stone at point A,u =10m/sec
Force constant of spring, k=2.0 N/m
Coefficient of kinetic friction between stone and horizontal ground =0.2
Distance covered by stone in rough horizontal region, L=100 m
Vertical distance between point A and point B,h=20m
Horizontal distance b/w points A & B,b=15 m No need to use this data in the problem
There is also no friction between point A and B.
Solution:
This problem will be solved using conservation energy.
B/W points A and B:
The total energy of stone at A=The total energy of stone in B = There’s no energy loss between these two points.
(Potential Energy)A +(Kinetic Energy)A =(Kinetic energy)B
=> mgh+(1/2)*m*(uA)2 =(1/2)*m*(vB)2
=> gh+(1/2)*(uA)2 =(1/2)*(vB)2—-(eq.i)
Where v B is the final velocity of stone at point A after it has come down from point A,
Gravitation =9.81 m/sec
This datum can be taken as horizontal ground. The potential energy of stone at point A=0 J
We get the following results by putting the values in eq (i).
9.81*20+(1/2)*102 =(1/2)*vB2
=> 196.2 J +50J =0.5*v 2
We solve for v and B.
vB=22.19 m/sec
Velocity at point after coming down from A on the hill.v B=22.19m/sec —[Ans [i]
Now, at maximum spring compression, the final velocity of stone,v =0 m/sec.
Stone is subject to frictional force
N is the normal reaction to stone, which is equal in weight.
N=mg=15*9.81 = N=147.15 Newton
F =29.43 Newton in reverse or opposite to the stone movement.
To overcome frictonal resistance, work should be done in stone.
W F =F *L Using work done=force displacement in force
W F =-29.43*100 joules -ve sign means that force applied is in the opposite direction to displacement.
So W f=2943J or N-m
Apply energy conservation again b/w point A and point where maximum spring compression is achieved (let point C).
Kinetic energy at point B = work done by stone to overcome frictonal resistance(Wf)+Potential energy of the spring
where x is maximum spring compression
Add the following values to the equation.
x=27.386m [Ans. (ii)]
Thank you. ***
Answer 3
Given
Mass m = 15kg
initial height hi = 20 m
a) From conservation of energy
1/2mvb^2 = mghi+1/2mva^2
vb = sqrt (va2+2ghi), sqrt (102+2*9.8*20), = 22.18m/s
b)
When the spring is compressed by the stone, it is at rest. Its final speed is zero
From the work energy theorem
Total work = chnage of kinetic energy
1/2kx^2-f*(100+x) = -1/2mvb^2
1/2*2*x^2-0.2*15*9.8(100+x) = -1/2*15*22.18^2
x^2 -2940-29.4x = 3690
x^2-29.4x-749.71 = 0
x = 16.38m
Answer 4
Conclusion
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