Question

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 1

-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-

KINDLY UPVOTE THE ANSWER IF YOU LIKED IT, IT WILL MAKE MY DAY ?

FOR ANY QUERIES COMMENT BELOW.

-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-

Answer 2

Given:

Mass of the stone,m=15 kg

Initial velocity of the stone i.e. velocity of stone at point A,u_A=10 m/sec

Force constant of the spring,k=2.0 N/m

Coefficient of kinetic friction b/w stone and the horizontal ground,=0.2

Distance covered by stone on rough horizontal region,L=100 m

Vertical distance b/w point A and B,h=20 m

Horizontal distance b/w point A & B,b=15 m {No use of this data in this problem}

Also,there is no friction on the hill b/w point A and point B.

Solution:

In this problem, we will use conservation of energy to solve.

B/W point A and B:

Total energy of stone at point A=Total energy of stone at point B {There is no loss of energy b/w 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 final velocity of stone at point B after coming down from point A of the hill,

and g is accelaraion due to gravity =9.81 m/sec²

Here datum is taken as horizontal ground,so potential energy of stone at point B=0 J

Putting the values in eq(i),we get

9.81*20+(1/2)*10² =(1/2)*v_B²

=> 196.2 J +50 J =0.5*v_B²

Solving for v_B, we get

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.

Frictional force acting on stone,

F_f=*N Newton

Where N is normal reaction on stone which is equal to the weight of stone.

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

So,F_f =*N=0.2*147.15

=> F_f=29.43 Newton in the backward direction or opposite to the stone motion.

So work done by stone to overcome frictonal resistance.

W_f =F_f*L {Using work done=force *displacement in the direction of force}

W_f =-29.43*100 Joule {-ve sign indicates that force applied and displacement are in opposite direction.So, energy is consumed in this process. }

So, W_f=2943 J or N-m

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

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

where x is the maximum compression of the spring.

Putting the values in the above eq.

So the maximum compression obtained in the spring when stone hits the spring,x=27.386 m [Ans (ii)]

***Thank you.If you found this solution helpful then please give a thumbs up and any feedback regarding solution in the comment box is appreciated.***

Answer 3

given

mass m = 15 kg

initial height hi = 20 m

a) from conservation of energy

1/2mvb^2 = mghi+1/2mva^2

vb = sqrt(va^2+2ghi) = sqrt(10^2+2*9.8*20) = 22.18 m/s

b)

when the stone compress the spring then it comes to rest that is final speed is zero

from work energy theorem

total work = chnage in 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 = 16.38 m

Answer 4