Block dropped onto a spring down an incline

[esoteric deviance]

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.

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/m​

h = ((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?

 

[OlderDan]

The units for your k are not N/m, so be careful what you do next. You have everything you need to find L and L is proportional to the change in height of the mass from the initial point to where it touches the spring. Use conservation of energy again to find the KE and the velocity.

 

[esoteric deviance]

Oh..yeeeah, meant to type N/cm sorry lol.

K, so I found L

L = (h/sin(30)) - x = 0.29 m​

(switched to meters b/c the answer in the book switched to meters, though I don't know why)​

and then used that to find the new h

h = Lsin(30) = 0.29sin(30) = 0.15 m​

which I then plugged into mgh = (1/2)mv^2 to find the velocity

v^2 = 2gh = 2(9.8)(0.15) = 2.94​

v = 1.7 m/s​

and I found the mgh = 1/2mv^2 by using the following

KE initial + PE initial = KE final + PE final​

0 + mgh = ((1/2)mv^2) + 0​

I know the answer is right, but did I find it the right way?
B/c someone was telling me that the h should = -Lsin(30) rather than +Lsin(30)...

 

[OlderDan]

The way you defined things, the + sign is correct. Someone else might have defined the zero of potental energy at the top and used negatives from there. Your approach and calculation are fine.

 

[esoteric deviance]

Ah, okay .


Thanks for the help ^_^.