A 2.40 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.40 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0380 m The spring has force constant 900 N/m The coefficient of kinetic friction between the floor and the block is 0.43 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 0.0150 m from its initial position? (At this point the spring is compressed 0.0230 m ) Express your answer with the appropriate units.

Explanation / Answer

apply the law of conservation of energy as

Initial potential energy of sprinf block system = final potential energy+kinetic energy of block+energy loss to friction

0.5 kx2^2 = 0.5 kx1^2 + 0.5 mv^2 + u mgh

i.e

0.5*900*0.038^2=0.5*900*0.023^2+0.5*2.4* V+0.43*2.4*9.8*0.015

Solving for v,

v=0.2167 m/s or 21.67cm/s

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