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

Question

A 2.40kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0330m . The spring has force constant 845N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . The block and spring are released from rest and the block slides along the floor.

What is the speed of the block when it has moved a distance of 0.0100m from its initial position? (At this point the spring is compressed 0.0230m .)

Can you please help me set up the problem? More interested in learning what to do than the answer.

Explanation / Answer

energy lost by spring = [kinetic energy gained by block] + [work done against friction]

energy lost by spring = 1/2 * k * ( x12 - x22 )

= 1/2 * 845 * (0.033^2 - 0.023^2)

= 0.2366 J

frictional force = mu * mg = 0.45*2.4*9.8 = 10.584

work done against friction = frictional force * distance

= 10.584 * 0.01 = 0.10584 J

substituting in energy Eqn
=> 0.2366 = 0.10584 + 1/2 * m * v2

=> v = 0.33 m/s

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