- #1
Reverend Lee
- 5
- 0
A 10kg box slides 4.0m down a frictionless ramp, then collides with a spring whose spring constant is 250 N/m. (Also, the ramp is angled at 30 degrees).
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.
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.