A 14.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leavin

Question

A 14.0 kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point A with 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 2.50 N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Part A

What is the speed of the stone when it reaches point B ?

Part B

How far will the stone compress the spring?

Explanation / Answer

potential energy of the stone at point A=mass*g*height=14*9.8*20=2744 J

kinetic energy of the stone=0.5*mass*speed^2=0.5*14*10^2=700 J

total mechanical energy of the stone=2744+700=3444 J

as there is no friction from A to B, energy conservation principle will hold true

at B, height=0==>potential enegy=0

let speed of the stone at B is v m/s

then kinetic energy=0.5*14*v^2=7*v^2

equating both initial and final energy:

7*v^2=3444

==>v=22.18 m/s

hence speed of the stone when it reaches point B is 22.18 m/s

part B:

let the stone compresses the spring by distance d m

then total kinetic energy of the stone will get converted into work done against friction and potential energy of the spring

friction force=kinetic friction coefficient*mass*g=0.2*14*9.8=27.44 N

work done against friction=force*distance=27.44*(d+100) J

potential energy in the spring=0.5*spring constant*compression^2

=0.5*2.5*d^2=1.25*d^2

hence equating kinetic energy of the stone at B to work done against friction and potential energy in the spring:

3444=27.44*(d+100)+1.25*d^2

==>1.25*d^2+27.44*d-700=0

==>d=15.11 m

hence the stone will compress the spring by 15.11 m.

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