3.5 kg box is launched from a spring that has spring constant 1200 N/m. The spring is initially compressed by an amount 1.7 m. The box slides along the frictionless track shown below and into a second spring that has a spring constant 900 N/m. It compresses the spring by an amount 2.1 m before turning around. What is the height of the second spring?
F=ma (for 3 different phases)
F=-kx (hookes equations)
V_f^2 = V_0^2 + 2a(delta_y)
The Attempt at a Solution
I feel like my attempt at a solution went too smoothly, so i'm suspicious I did something terribly wrong somewhere, or forgot to account for something.
Basically I started at the blocks starting position. Using Hooke's law, I got F=-kx, and F turns out to be -2040 N... (-1200N/m / 1.7m). I then do the same thing for the other side where the spring compresses at the end of the blocks path. I get F= -1890N ... (-900N/M)(2.1m).
So there is 150 N lost in the system... this must mean 150N is lost when its going up along the higher slope on the other side?
I then attempt to find V_final using mgh = (1/2)mv_f^2 - (1/2)mv_i^2. But v_i^2 is just 0 so...
I plug in numbers, do the algebra and v_f comes out to be 11.46 m/s ...
I THEN use V_f^2 = V_0^2 + 2a(delta_y)
Same story... plug in numbers and do algebra and I get delta_y to be 1.53m.
Now I tack on the starting height plus this found height. So I get 6.7m + 1.53m and I end up with the final answer of 8.23m
Does this all sound right?
I dont have a solution to the problem so I wanted to check with the pro's on this forum.
Thanks for any help!