20.0m above the ground, the cable brakes and the bricks begin to fall. If there is a safety net set up 2.00m about the ground, what does the spring constant of the net have to be so that the crate of bricks just touches the ground when the net stretches downward?
We have ke = mg(H - 2) = 1/2 kdX^2 = pe; so k = 2mg(H - h)/dX^2 = 2*500*9.8*(18)/4 = 44100 N/m ANS.
If the spring was not used, all of the crate’s potential energy would be converted into kinetic energy. For the crate to stop moving as the spring is compressed two meters, the work that is done by the spring must be equal to the initial potential energy of the crate. PE = 500 * 9.8 * 20 = 98,000 J Work = ? * k * d^2 = 2 * k 2 * k = 98,000 k = 49,000 N/m As the spring is compressed 2 meters, the velocity of the crate of bricks must decrease to 0 m/s. For this to happen, the work that is done by the spring must be equal the maximum kinetic energy of the crate. To determine the velocity of the crate of bricks if they were to fall 18 meters, use the following equation. vf^2 = vi^2 + 2 * a * d, vi = 0, a = 9.8, d = 18 vf^2 = 2 * 9.8 * 18 = 352.8 vf = √352.8 This is approximately 18.8 m/s. KE = ? * 500 * 352.8= 88,200 As the crate move 2 meters downward, it potential energy increases. PE = 500 * 9.8 * 2 = 9800 J Total energy = 88,200 + 9800 = 98,000 J Work = ? * k * d^2, d = 2 meter Work = ? * k * 2^2 = 2 * k 2 * k = 98,000 k = 49,000 N/m This is exactly the same answer as above.
