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A 1.50-kg block is on a frictionless, 30 degrees inclined plane. The block is attached to...

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Physics

A 1.50-kg block is on a frictionless, 30 degrees inclined plane.The block is attached to a spring (k = 40.0N/m ) that isfixed to a wall at the bottom of the incline. A light stringattached to the block runs over a frictionless pulley to a 60.0-gsuspended mass. The suspended mass is given an initial downwardspeed of 1.40m/s. How far does it drop before coming to rest?(Assume the spring is unlimited in how far it can stretch.)

y = _____ m

Please show steps I am confused.

Answer & Explanation Solved by verified expert
4.4 Ratings (706 Votes)
Solution Let M1 blocks mass M2 suspended mass theta angle of incline M1 150 kg M2 600 g 600 103 kg 006 kg theta 30 deg Forces on M2 are 1 Weight M2 g downward 2 Tension T in the string upward Initially M2 is in equilibrium therefore net force on it 0 Therefore T M2 g Or T 006 98 Or T 0588 N1 Let x1 length by which the spring is initially compressed Forces on the block parallel to the incline are 1 Force kx1 40 x1 by spring up the incline 2 Component of weight M1g sintheta 150 98 sin30 deg 735 N down the incline 3 Tension T up the incline The block is    See Answer
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