- #1

Reverend Lee

- 5

- 0

a). What is the maximum compression of the spring?

b). At what compression of the spring does the box have its maximum velocity?

I started the problem by finding the final velocity of the block at the end of the 4m distance (which is the beginning of the spring) by finding the acceleration ( a=9.8*sin(30) ), then I applied it to the equation Vf ^2= Vi ^2 * 2a(delta x), and received an answer of Vf=6.261 m/s. I used this result for the initial velocity of the block moving down on the spring. After that, I used the equation

.5*m*(Vf^2)+.5*k*(Xf - Xequilibrium)^2 +mgXf =.5m(Vi^2) + .5k(Xi - Xequilibrium)^2 + mgXi

I considered Xi = Xequilibrium = 0; Vf = 0; and Vi =6.261

but got the wrong answer (0.88m). The correct answer is 1.46m. I'm not really sure how to approach part b either.