A 2.60 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.60 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0310 m . The spring has force constant 865 N/m . The coefficient of kinetic friction between the floor and the block is 0.39 . 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 0.0190 m from its initial position? (At this point the spring is compressed 0.0120 m .) Express your answer with the appropriate units.

Explanation / Answer

Initially the spring energy is U = ½kx² = ½ * 865N/m * (0.0310m)² = 0.416 J
Later the spring energy is U = ½ * 865N/m * (0.0120m)² = 0.062 J
and the work done W = µmgd = 0.39 * 2.6kg * 9.8m/s² * 0.0190m = 0.189 J

KE = U - U - W = (0.416 - 0.062 - 0.189) J = 0.165 J
0.165 J = ½mv² = ½ * 2.6kg * v²
v = 0.356 m/s

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