**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 =10 m/sec

Force constant of spring, k=2.0 N/m

Coefficient of kinetic friction b/w stone and the 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 This data is not used in this problem.

There is also no friction between point A and B.

**Solution:**

This problem will be solved using **conservation of
energy**

**B/W point 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*(u_{A})^{2} =(1/2)*m*(v_{B})^{2}

=> gh+(1/2)*(u_{A})^{2} =(1/2)*(v_{B})^{2}**—-(eq.i)**

Where v _{B} is the final velocity of stone at point A after it has come down from the top of the hill.

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)*10^{2} =(1/2)*v_{B}^{2}

=> 196.2 J +50J =0.5*v 2

We solve for v _{and B}.

v_{B}=22.19 m/sec

**Velocity
of stone at point after coming down from point A of the
hill,v**

_{B}=22.19 m/sec —[Ans (i)]

Now, **at maximum compression of the spring,the final
velocity of stone,v _{f}=0 m/sec.**

Stone is subject to frictional force

*N Newton

N is the normal reaction to stone, which is equal in weight.

i.e.N=mg=15*9.81=> N=147.15 Newton

*N=0.2*147.15

=> **F _{f}=29.43 Newton in the backward
direction or opposite to the stone motion.**

To overcome frictonal resistance, work should be done in stone.

W _{F} =F *L Using work done=force displacement in force direction

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 **conservation of energy b/w point B and
point where maximum compression of spring is obtained(let point
C)** again

Kinetic energy at point B = work done by stone to overcome frictonal resistance(W_{f})+Potential energy of the spring

where x is maximum spring compression

Add the following values to the equation.

The maximum spring compression at the point where stone touches the spring is $strongTag10

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 stone compresses the spring, it comes to 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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