A 2.20 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.20 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0380 m . The spring has force constant 860 N/m . The coefficient of kinetic friction between the floor and the block is 0.42 . The block and spring are released from rest and the block slides along the floor.

Part A

What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0180 m .)

Express your answer with the appropriate units.

Explanation / Answer

friction force=friction coefficient*normal force

=friction coefficient*weight of the block

=0.42*2.2*9.8

=9.0552 N

as the block is moving on a level horizontal surface, there is no change in its potential energy

hence it will not be included in the work energy principle equation.

let final speed of the block is v m/s.
using work energy principle:

initial potential energy of the spring + initial kinetic energy of the block - work done against friction

=final potential energy of the spring + final kinetic energy of the spring

==>0.5*spring constant*compression^2+0.5*mass*initial speed^2-friction force*distance covered

=0.5*spring constant*final comprression^2+0.5*mass*final speed^2

==>0.5*860*0.038^2+0.5*2.2*0^2-9.0552*0.02=0.5*860*0.018^2+0.5*2.2*v^2

==>v=sqrt((0.5*860*0.038^2+0.5*2.2*0^2-9.0552*0.02-0.5*860*0.018^2)/(0.5*2.2))

=0.52266 m/s

hence speed of the block when it has moved a distance of 0.02 m is 0.52266 m/s

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