I tried using the equation U1=K2, or mgh=1/2kx^2, but it didn't work. I'm confused about how to find the height and where the angle comes into the problem, and how to figure it out?
You are definitely on the right track for this problem. It is a conservation of energy problem.
You are calling position 1 the position where the block is at the top of the incline. Since the block is at rest at the top of the incline, K1=0. Since the spring is not compressed or stretched at that position, U(spring)=0 too. You are calling position 2 the position where the block has come to rest and the spring is fully compressed. at that point the U2 (gravitational potential energy) is zero. However, the block is at rest there, so the K2 is also zero. But the spring is fully compressed so the U(spring) is not zero, and it is (1/2)kx^2.
So your equation is really Ug1+Us1+K=Ug2+Us2+K, and when you put in the zeros it becomes Ug1=Us2. And that can be written as mgh=(1/2)kx^2.
OK, so how to find the height. The problem gives you the length of the incline and its angle (from the horizontal, the way we usually draw inclines). If you draw the incline as a triangle, with the incline the hypotenuse (the way we usually draw inclines), you can use the angle and the hypotenuse (the length of the incline) to find the height of the incline. That's your h.
2 comments:
You are definitely on the right track for this problem. It is a conservation of energy problem.
You are calling position 1 the position where the block is at the top of the incline. Since the block is at rest at the top of the incline, K1=0. Since the spring is not compressed or stretched at that position, U(spring)=0 too. You are calling position 2 the position where the block has come to rest and the spring is fully compressed. at that point the U2 (gravitational potential energy) is zero. However, the block is at rest there, so the K2 is also zero. But the spring is fully compressed so the U(spring) is not zero, and it is (1/2)kx^2.
So your equation is really Ug1+Us1+K=Ug2+Us2+K, and when you put in the zeros it becomes Ug1=Us2. And that can be written as mgh=(1/2)kx^2.
OK, so how to find the height. The problem gives you the length of the incline and its angle (from the horizontal, the way we usually draw inclines). If you draw the incline as a triangle, with the incline the hypotenuse (the way we usually draw inclines), you can use the angle and the hypotenuse (the length of the incline) to find the height of the incline. That's your h.
Hope that helps. Good luck.
Got it, thank you!
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