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A 1.5 kg block is attached to a spring with a spring constant of 1950 N/m. The spring is then stretched a distance of 0.33 cm and the block is released from rest.

(a) Calculate the speed of the block as it passes through the equilibrium position if no friction is present.

(b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 2.0 N retards its motion.

(c) What would be the strength of the frictional force if the block reached the equilibrium position the first time with zero velocity?

I got part (a) by by saying that 1/2mv^2=1/2kA^2 and my answer of 11.898 cm/s was correct. but i can't get (b) or (c). in (b) i said that the work done by friction (2*0.33) was equal to 1/2mv^2 - 1/2kA^2, but my answer is incorrect.

i have no clue how to do (c). please help me!

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# Homework Help: Spring equilibrium

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