Darth Maul examines a 1.0 kg object that is moving along a horizontal surface (w

Question

Darth Maul examines a 1.0 kg object that is moving along a horizontal surface (where the coefficient of kinetic friction µk = 0.30. The object has a speed v = 1.0 m/s when it hits a massless spring head-on (as in the figure).

(a) If the spring has a force constant k = 120 N/m, how far will Darth Maul see the spring compress?
  m
Hint: Darth Maul has been known to make use of the quadratic equation when solving problems at times ... it will be necessary for this problem as well!

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

Hint: When the spring is fully compressed, how much force is the spring applying to the object? When will it be enough to get the object to actually move?

(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).]
m/s
Hint: Be careful with rounding errors. Write down more digits than you need ... and use those "longer" numbers for future calculations (not the rounded numbers!).

Explanation / Answer

(a) By the law of energy conservation:-
=>KE(initial) = PE(spring) + W(friction)
=>1/2mu^2 = 1/2kx^2 + Ff.x
=>1/2 x 1 x (1)^2 = 1/2 x 120 x (x)^2 + µ x mg x (x)
=>1/2 x 1 x (1)^2 = 1/2 x 120 x (x)^2 + 0.30 x 1 x 9.8 x (x)
=>60(x)^2 + 2.94(x) - .5 = 0
=>(x) = 0.07 m or 7 cm
(b) By F(friction) = F(applied by spring)
=>µs x mg = kx
=>µs = [120 x 0.07]/[1 x 9.8]
=>µs = 0.86
(c) By the law of energy conservation:-
=>KE(final) = PE(spring) - W(friction)
=>1/2mv^2 = 1/2kx^2 - Ff .x
=>1/2 x 1 x v^2 = 1/2 x 120 x (0.07)^2 - 0.3 x 1 x 9.8 x 0.07
=>v^2 = 0.588 - 0.2058
v=0.62 m/s

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