Recent content by KAT444

I agree with your profile, you're an all around good guy. Thanks again - KAT444

Is this the right way: elastic PE = 1/2kx^2 energy is conserved so . . . 9810J = 1/2(1000N/m)(x^2) 4.43m = x Thanks - KAT444

Is this right? : 1/2mv^2 + mgy = 1/2mv^2 + mgy 0 + (50kg)(9.81m/s^2)(20m) = 1/2(50kg)v^2 + 0 19.81m/s = v So then the spring would be compressed .9905m because: f = ma f = 50kg(19.81m/s) f = 990.5N fp = kx 990.5N = 1000N/m * x .9905m = x Thanks for all your help -KAT444

I don't want you to give me the answer, but is there anyway you could give me a little more information? I know that energy should be conserved, but the only formulas I know for the conservation of energy involve velocity, which I can't figure out a way to find.

[SOLVED] spring and box on incline Homework Statement A 50 kg mass rests atop a 20m high incline. The spring constant of a 10m long spring below is 1000N/m. The 50 kg crate is sent down the incline. Assume all surfaces in this problem are frictionless. What is the velocity of the crate...