Calculate Change in Kinetic Energy for Masses on Pulley

[asianguy361]

Two masses are connected by a string via a pulley, m1=45kg is on a ramp while m2=103kg is hanging off the edge. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 45.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 45.0 kg block as it moves from A to B, a distance of 20.0 m. Theta is 37 degrees


I first tried to find velocity (V) to use in a DKE equation
D=delta

My equation for V was sqrt((2(m2ghi-m1ghf+um1gdcos@))/(m1+m2))

u = coefficient of kinetic friction
hi = initial height, 20m in this case
hf = 20sin@
@ = theta

I then used DKE = 0.5m1Vf^2-0.5m1Vi^2

where Vi = 0

Using this method I got an answer of DKE of 5059.76J, which was wrong.
I do not know if I am over simplifying the problem or thinking too much or if I'm completely thinking in the wrong direction. Help would be much appreciated.

 

[Doc Al]

My equation for V was sqrt((2(m2ghi-m1ghf+um1gdcos@))/(m1+m2))

Check the signs of your terms.

 

[Moetasim]

I understand your thought process and appreciate your efforts in trying to solve this problem. However, I believe your approach may be overcomplicating the problem and may not be taking into account all the forces involved. Let me explain how I would approach this problem.

First, let's draw a free-body diagram for the 45kg block on the ramp. We have the force of gravity (mg) acting downward, the normal force (N) acting perpendicular to the ramp, and the force of friction (F) acting parallel to the ramp. Since the block is moving, we also have a net force (Fnet) acting in the direction of motion.

Next, let's consider the forces acting on the 103kg block hanging off the edge. We have the force of gravity (mg) acting downward, and the tension in the string (T) acting upward.

Since the pulley is frictionless and of negligible mass, we can assume that the tension in the string is the same on both sides of the pulley. This means that T = Fnet.

Now, let's consider the motion of the 45kg block. As it moves from A to B, its initial kinetic energy (KEi) is zero, and its final kinetic energy (KEf) is given by KEf = 0.5m1Vf^2.

To find the final velocity (Vf), we can use the work-energy theorem, which states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. In this case, the work done by the net force (Fnet) is equal to the change in kinetic energy of the 45kg block.

So, we can write the work-energy equation as:

W = Fnet * d = KEf - KEi

Where W is the work done by the net force, d is the distance traveled, and KEi is the initial kinetic energy (which we have assumed to be zero).

Now, let's calculate the work done by the net force. We can break this down into the work done by gravity (Wg) and the work done by friction (Wf).

Wg = m1ghf - m1ghi

Where hi is the initial height (20m) and hf is the final height (20sin37 = 12.6m).

Wf = F * d = um1gdcos@ * d