A 2.50 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.50 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0330 m . The spring has force constant 845 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.

What is the speed of the block when it has moved a distance of 0.0110 m from its initial position? (At this point the spring is compressed 0.0220 m .)

Express your answer with the appropriate units.

Explanation / Answer

the

initial spring energy = final energy + KE of the block + the work done by the friction

0.5 X 845 X 0.0332 = 0.5 X 845 X 0.0222 + 0.5 X 2.5 X v2 + 2.5 X 9.8 X 0.42 X 0.011

0.4601 = 0.2044 + 1.25 v2 + 0.1131

1.25 v2 = 0.1426

v2 = 0.11408

v = 0.3377 m/sec

the speed of the block when it has moved a distance of 0.0110 m from its initial position is v = 0.3377 m/sec

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