Isn't this suppose to be easy can't get it :(

Question

anonymous · Answer

attachment

anonymous · Answer

yea mgh/costheta=1/2 kx^2

anonymous · Answer

I don't know the length of the spring tho...

anonymous · Answer

the block doesn't start from the top which is 3 m

anonymous · Answer

so i can't seem to use h as 3...

anonymous · Answer

no it's released from the top. you can use the Newton's second law to obtain the distance traveled by the object.

anonymous · Answer

really so i assume that it's released from the very top ?

anonymous · Answer

what's with the way they worded it with the spring not being compressed or stretched wouldn't i have to assume some distance say (5-n) for the ramp distance? in which that is the initial position?

anonymous · Answer

I think that it is not too hard you are imagining. write the second law statement and obtain the distance x that the spring has elongated.

mg sin theta = kx --> x = ...

anonymous · Answer

ohh wow im how did i not see that Iuuno why im trying to overcomplicate things here it's just simply a question where Force applied =spring force .... silly me