(a) If the spring has force constant k = 120 N/m, how far is the spring compress

Question

(a) If the spring has force constant k= 120 N/m, how far is the spring compressed?
1 m

(b) What minimum value of the coefficient of static friction,µs, will assure that the spring remains compressedat the maximum compressed position?
2

(c) If µs is less than this, what is the speed ofthe block when it detaches from the decompressing spring? [Hint:Detachment occurs when the spring reaches its natural length(x = 0).]
3 m/s

Explain why detachment occurs when the spring reaches its naturallength.
4 4

Explanation / Answer

A) Using the conservation of energy: mv2/2 = kmgx +kx2/2 2(1.5)2/2 = 0.30(2) ( 9.80) x + 120x2/2: this is a quadratic equation and we solve for x. => x = 0.151m, -0.249m, where we ignore the negativeresult. B) smg = kx=>  s= kx/mg = 120(0.151)/2(9.80) =0.924 C) Again from the conservation of energy: mv'2/2 = mv2/2 -2kmgxmax = > v' =(2.25-1.77576)1/2 = 0.6886m/s D) When the spring reaches its natural position it experiencesno force and hence it does not stretches further.

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