Recent content by marky

Sorry if you read my original first solution. It was wrong. I forgot to include the work done by gravity. Anyways, I made the changes. There are two approaches. Both of which have the same answer. The second seems shorter. But often using work and energy can simplify things which is why I...

Yes, sure. Applying v_f^2 = v_i^2 + 2ad Let's look at the point from x = 0 to x=0.8. This is where the applied force is being exerted. Your FBD should have the force of gravity and the applied force. Summing F = ma you get a = 8.5 m/s^2. v_i = 0 v_f = \sqrt{ 2 (8.5)(0.8)} = 3.688 m/s...

EDIT: Forgot to include work done by gravity. You have a force acting over a distance. That means work is done on the system. Work done, U = Fd - mgh = (55 N)*(.8m) - (0.8m)(9.81)(3) = 20.5 J Work and Energy Equation T_i + U = T_f T is kinetic energy T = \frac{1}{2}mv^2 T_i...