A block of mass m = 12 kg is released from rest on a frictionless incline of angle theta = 30 degrees. Below the block is a spring that can be compressed 2.0 cm by a force of 270 N. The block momentarily stops when it compresses the spring by 5.5 cm.(adsbygoogle = window.adsbygoogle || []).push({});

a) How far does the block move down the incline from its rest position to this stopping point?

b) What is the speed of the block just as it touches the spring?

Okay, so I got the answer to part a like this:

k = F/x = 270/2.0 = 135 N/mh = ((1/2)kx^2)/(mg) = 17.4 cm (used law of conservation of energy for this one)(L + x) = h/sin(30) = 35 cm

But i'm having some trouble with part b .

Anyone think they can help me out a bit with part b?

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# Homework Help: Block dropped onto a spring down an incline

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