A 5.40 kg block free to move on a horizontal, frictionless surface is attached t

Question

A 5.40 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 0.122 m from equilibrium and is then released. The speed of the block is 1.40 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 mu_k = 0.285. Determine the speed of the block at the equilibrium position of the spring.

Explanation / Answer

Applying work - energy theorem,

Work done by spring force = change in KE

k x^2 /2 = m v0^2 /2 -0

k(0.122^2) /2 = 5.40(1.40^2)/2

k = 711.1 N/m

when friction force is acting then fk = uk N = uk m g = 0.285 x 5.40 x 9.8 = 15.08 N

Now applying work- energy theorem,

Work done by spring + work done by spring = change in KE

711.1 ( 0.122^2)/2 + ( - 15.08 x 0.122) = m v^2 / 2 - 0

5.292 - 1.84 = 5.40 v^2 /2

v = 1.13 m/s

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