5. You release a block with a mass 2.00 kg on an incline plane. The box is relea

Question

5. You release a block with a mass 2.00 kg on an incline plane. The box is released 4.00 m from a long spring with a spring force constant of 120 N/m, that is attached at the bottom of the incline. The 53.19 incline is not frictionless and the coefficient of friction between the block and the incline are 0.40 and -2.00 kg 4.00 m 0.20. For this problem, you can consider that the mass of the spring is negligible. 53.1° a) What is the speed of the block just before it reaches the spring? b) What is the maximum compression of the spring? ) After reaching the bottom, the block rebounds back up the incline. How close does it get to its initial position?

Explanation / Answer

5. (A) Work - energy theorem.

Work done by gravity + work done by friction = change in KE

m g L sin53.1 - fk. L = m v^2 /2 - 0

m g L sin53.1 - uk m g L cos53.1 = m v^2 / 2

v^2 = 2 g L ( sin53.1 - uk cos53.1)

v^2 = (2 x 9.8 x 4) (sin53.1 - (0.20 x cos53.1))

v = 7.30 m/s

(B) Applying work - energy theorem,

m g d sin53.1 - uk m g d cos53.1 - k d^2 / 2 = 0 - m v^2 /2

(2 x 9.8 x d x sin53.1) - (0.2 x 2 x 9.8 x d x cos53.1) - (120 d^2 / 2) = 0 - (2 x 7.30^2 / 2)

15.67d - 2.35d - 60 d^2 = - 53.28

60 d^2 - 13.32 d - 53.28 = 0

d = 1.06 m

(D) - (2 x 9.8 x (1.06 + d) sin53.1) - (0.2 x 2 x 9.8 x (1.06 + d) cos53.1) + (120 x 1.06^2 / 2) = 0 - 0

- 16.614 - 15.67d - 2.49 - 2.35 d + 67.416 = 0

d = 2.68 m

close to initial position = 4 - 2.68 = 1.32 m

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