Question

A 5 kg mass is thrown up with a velocity of 4 m/s from a height of 30 m onto a spring with a relaxed length of 10 m, and a constant of 400 N/m. What will the maximum compression of the spring be? What will the speed of the object be when the spring is compressed 0.3 m?

HERE'S MY WORK:

(V2^2)=(Vi)^2+2ad
0=16-19.6d
-16=-19.6d
d=0.816 m + 30 m = 30.816 meters above the ground

mgh = mgh + 1/2kx^2 = (5 kg)(9.8 m/s^2)(30.816) = 200x^2-49x+490

x=2.75 m

mgh=1/2kx^2+mgh+1/2mv^2
(5)(9.8)(30.186)=1/2(400)(0.3)^2+(5)(9.8)(9.7)+1/2(5v^2)

v=20.17 m/s

Is it correct?

Answer 1

yes.

Answer 2

I plugged in my value for x, but both sides of the equation had different values, so I think I have the wrong value for x. How do I solve 200x^2-49x+490?

Answer 3

use the quadratic equation. I didn't look at your work, but used energy relationships (Initial KE+ initial PE=springenergy- mgx where x is the compression. That was the same as your x, 2.75m ...if I didnt'make an error.