A 2.0 kg block is placed against a compressed spring on a 30 degree incline. The

Question

A 2.0 kg block is placed against a compressed spring on a 30 degree incline. The spring whose force constant is 19.6 N/m, is compressed 20 cm, after which the block is released and comes to rest at a distance up the incline. The coefficient of kinetic friction is 0.10. Write out the conservation of energy equation including the known data from the problem statement. Set y = 0 to be the initial height of the block before it is release Calculate how far up the incline the block will it go before coming to rest? final position with respect to its position just before being released.

Explanation / Answer

both answers are below

work done by all forces = change in K.E.

1/2 KX^2 + [- mg (d)] + mg sin (30) d = 0

1/2 x 19.6 x (20 x 10^-2)^2 - 2 x 9.8 x .5 x d = .10 x 2 x 9.8 x d

d = .03 m

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