Question

13.2)

A 1.3 -kg block slides along a horizontal surface with a coefficient of friction μk = 0.274. The block has a speed v = 2.28 m/s when it strikes a massless spring head-on.

a. If the spring has a force constant k = 30.9 N/m, how far is the spring compressed?

b. What minimum value of the coefficient of static friction, μs, will assure the spring remains compressed at the maximum compressed position?

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

Answer 1

a)

using conservation of energy

0.5 m v^2 - u mg x = 0.5 k x^2

0.5* 1.3* 2.28^2- 0.274* 1.3* 9.8 x = 15.45 x^2

15.45 x^2 + 3.491 x - 3.379 = 0... (i)

solving for x

x = 0.3681 m

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b)

using hooke's law

u mg = k x

u = 30.9* 0.3681 / (1.3* 9.8)

u = 0.8928

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c)

0.5 k x^2 =0.5 mv^2

30.9* 0.3681^2 = 1.3* v^2

v = 1.7946 m/s

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