A 1.50 kg block collides with a horizontal weightless spring of force constant 7.70 N/m
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A 1.50 kg block collides with a horizontal weightless spring of force constant 7.70 N/m

[From: ] [author: ] [Date: 11-11-10] [Hit: ]
5kg x 0.25 times the distance covered. add the two energies, then plug them into the kinetic energy formula and then you work out the velocity.......
The block compresses the spring 4.00 m from the rest position. Assuming that the coefficient of kinetic friction between the block and the horizontal surface is 0.250 , what was the speed of the block (in meters/second) at the instant of collision?

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you find the potential energy gained with the compression of the spring which is 1/2 k (d)square, the work done by the resisting force which is 1.5kg x 0.25 times the distance covered. add the two energies, then plug them into the kinetic energy formula and then you work out the velocity. too tired to do the calculations for you
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