A moving 1.90 kg block collides with a horizontal spring whose spring constant i

Question

A moving 1.90 kg block collides with a horizontal spring whose spring constant is 420 N/m. The block compresses the spring a maximum distance of 10.00 from its rest position. The coefficient of kinetic friction between block and the horizontal surface is 0.200. what is the work done the spring in bringing the block to rest? How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? what is the speed of the block when it hits the spring?

Explanation / Answer

1.
Work done by spring = ½ * Force * distance
Force = Spring constant * distance
Work done by spring = ½ * Spring Constant * distance ^2
Work done by spring = ½ * 0.2 * 0.10 ^2
Work done by spring = 1x10^-3 J

Work done by friction = µ * m *g * d
Work done by friction = 0.20 * 1.9 *9.8 * 0.10
=0.3724 J

When the block hits the spring, its Kinetic = Work done by the spring.
Ke = ½ * mass * velocity^2
Work done by spring = 1x10^-3 J
½ * mass * velocity^2 = 1x10^-3 J
½ * 1.9 * velocity^2 = 1x10^-3 J
velocity = .0324 m/s


Velocity when block has compressed the spring 0.10m =
Kinetic energy when block compressed the spring 0.11m = –work done by friction -Work done by spring
KE =0.3724- 1x10^-3 =-.3714
½ * 1.9 * velocity^2 = .3714 J

V = 0.625 m/s

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