:( Please help! Work,power and energy question 1. The problem statement, all variables and given/known data There is a 100kg object on a 30-degree incline ramp. The mew of friction is 0.5. There is a motor on top of the ramp, which accelerates the object at .22 m/s^2 by a rope. Find the power output of the motor. 2. Relevant equations d=(vit)+(.5)(a)(t)^2 W=Fd F=ma P=w/t, E/s 3. The attempt at a solution I think I'm supposed to use Ek+Ep+TE=0, but how am I supposed to find the power without knowing the time? So, I've tried this method: P=J/s J(work)= Fxd So I thought maybe I'll use t=1: vi=0 vf=x a= 0.22 d=? t=1 d=vit+(.5)(a)(t)^2 d=(.5)(.22)(1) d=.11m Fnet=ma Fa-Ff=ma Fa-(100kg)(9.81)(cos30)(.5[mew])-(100kg)(9.81)(sin30)=(100)(.22) Fa-424.79-490.5=22 Fa=937.29N W=Fd W=(937.29)(.11) W=103.10J P=J/s P=103.10Watts So here's the one approach I had, ------------------ Here is the second Ek+Ep+TE=0 (.5)(m)(vf^2-vi^2) + mgh + (Ff)d=0 To find Vf, Vi=0 Vf=? a=0.22 d=x t=1s a=(Vf-Vi)/t .22=Vf/1 Vf=.22 m/s plugging it back in, (.5)(100kg)(0.0484) + (100)(9.81)(h) + (100)(9.81)(cos30)(0.5[mew])d=0 To find d, Vi=0 Vf=x a=.22 d=? t=1s d=vit+(.5)(a)(t^2) done above, d=0.11m (.5)(100kg)(0.0484) + (100)(9.81)(h) + (100)(9.81)(cos30)(0.5)(0.11) = 0 2.42 + 981h + 46.73 = 0 h=.05m Then plug the height of .05m, to get 98.30J/s? -------------- Another question: If there is an elevator moving downwards to the ground at a constant velocity, Is the work done 0J? Is the net work between Ft and Fg the same? How should I approach this question? I know there is NOT an acceleration. W=Fd, W=mad, so no work? Although there is a ΔEnergy? Another question: Momentum question, Is the "internal work" done by an explosion equal to the combined kinetic energy of the particles? Or is it 0J? How can i visualize internal work in a bomb-like scenario? Thank you very much.