A 5.43 kg block free to move on a horizontal, frictionless surface is attached t

Question

A 5.43 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is fixed. The spring is compressed 0.149 m from equilibrium and is then released. The speed of the block is 1.19 m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which ?k = 0.255. Determine the speed of the block at the equilibrium position of the spring.

Explanation / Answer

The difference is that now friction will do negative work against the block. Wf= umgx, Wf must be subtracted from the kinetic energy the block had at equilibrium in the absence of friction. KE = 0.5*mv^2. With friction, KE= 0.5*mv^2 - umgx, and this kinetic energy will be equal to 0.5*m(Vf)^2, where Vfis the velocity under friction. If you write 0.5*mv^2 - umgx = 0.5*m(Vf)^2, m cancels out, giving you v^2 - 2ugx = (Vf)^2. You can rewrite this as Vf = sqrt(v^2 - 2ugx). So Vf = sqrt(1.19^2 -2(0.255)(9.8)(0.149m) = 0.819m/s. Hope this helps!

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