A block of mass 0.5 kg is pushed against a hor- izontal spring of negligible mas

Question

A block of mass 0.5 kg is pushed against a hor- izontal spring of negligible mass, compressing the spring a distance of ?x as shown in the fig- ure. The spring constant is 458 N/m. When released, the block travels along a frictionless, horizontal surface to point B, the bottom of a vertical circular track of radius 1 m, and continues to move up the track. The speed of the block at the bottom of the track is 13 m/s, and the block experiences an aver- age frictional force of 7 N while sliding up the track. The acceleration of gravity is 9.8 m/s2 . What is the speed of the block at the top of the track? Answer in units of m/s

Explanation / Answer

we know the energy stored due to compression of the spring=( 1/2)*k*(x)^2

This energy is fully converted to kinetic energy as there are no damping or resistive forces.(By Conservation of energy)

so 1/2*m*v^2=( 1/2)*k*(x)^2 where v=speed of the block at the bottom of the track is 13 m/s

so, sqrt(m*v^2/k)=x=0.43m

along the track, the frictional force does work on the body and the body loses energy.

so By conservation of energy,

energy of the body at the bottom of the track,E=energy lost due to friction+potential energy gained from point B to T+ kinetic energy remaining

energy lost due to friction,Ef=F.s  

where F=avg frictional force on the body=7N,

s=distance travelled=circumference of the track=2*r=2*3.14*1=6.28m

so Ef=7*6.28=43.96J

potential energy gained from point B to T,U=mg*2r=0.5*9.8*2*1=9.8J

kinetic energy remaining=(1/2)*m*vt^2

so,E=1/2*0.5*13^2=Ef+U+(1/2)*m*vt^2

ie,   169=43.96+9.8+(.5*2*vt^2)

so    vt =10.73 m/s

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