A moving 3.10 kg block collides with a horizontal spring whose spring constant i

Question

A moving 3.10 kg block collides with a horizontal spring whose spring constant is 392 N/m. The block compresses the spring a maximum distance of 11.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. What is the work done by 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? So, three parts please.

Explanation / Answer

A) W spring = d PE spring = 1/2 kx^2 = 0.5*392*.11^2=2.37 J B) W friction= f d = u mg d = 0.25*3.1*9.81*.11=0.836 C) all kinetic energy went into friction and spring so 1/2 mv^2 = W spring + W friction 0.5*3.1*v^2=2.37+0.836 v=1.44 m/s

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