Solving 2 Weights on a Pulley Problem

[jfleury45]

1. A 3kg weight is attached to a 2kg weight by a rope. This rope is placed over a pully so the weights are hanging. The 2kg weight starts 4 meters lower than the 3kg weight. If the system starts at rest what is the speed of the objects when they are at the same height.



2. Fnet = ma
v^2=2*a*(x2-x1)




3. I took the net force to be the normal force of the 3kg weight minus the normal force of the 2 kg weight. Fnet=9.81N I then divided by 3kg to get a=3.27 m/s^2 I then plugged knowns into v^2=2*a*(x2-x1) v^2=2*3.27*2 and got v=3.617 This answer does not work. What am i doing wrong?

 

[robb_]

The normal force usually refers to a force provided by contact between the system and the environment. Try drawing a free body diagram for both objects, then write down Newtons second law for both objects. Think about the acceleration of each object, then solve for the acceleration of either. Plug into the kinematic equations.

 

[m2003]

I would first commend your attempt to use the principles of physics to solve this problem. However, there are a few things that could be causing your answer to be incorrect.

Firstly, it seems that you have used the wrong value for the acceleration due to gravity (g). In your calculation, you have used 9.81N as the net force, which is actually the value of g. The correct value to use for the acceleration in this problem would be 9.81 m/s^2.

Secondly, it appears that you have used the formula v^2=2*a*(x2-x1) incorrectly. This formula is used to calculate the final velocity (v) when an object is moving with a constant acceleration (a) over a distance (x2-x1). In this problem, the objects are starting from rest and will reach the same height, so the final velocity will be zero. Therefore, the formula should be v=√(2*a*(x2-x1)).

Finally, it is important to consider the direction of the forces and velocities in this problem. Since the 2kg weight is starting 4 meters lower than the 3kg weight, it will have a greater potential energy and therefore a greater downward force than the 3kg weight. This means that the net force should be the weight of the 3kg weight minus the weight of the 2kg weight, which is 3kg*9.81m/s^2 - 2kg*9.81m/s^2 = 19.62N.

By using the correct values and formula, the final velocity should be v=√(2*(19.62N/3kg)*(4m)), which gives a final velocity of 5.19 m/s.

In summary, to solve this problem correctly, you should use the correct value for g, use the correct formula for calculating final velocity, and consider the direction of forces and velocities in your calculations. I hope this helps you to solve the problem accurately.