Help with Newtons Law Pulley Problem

[ludakrishna]

1. Two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of 412N, and the other has a weight of 908N. The rope and the pulleys are massless and there is no friction. a.) What is the acceleration of the lighter block? b.) Suppose that the heavier block is removed and a downward force of 908N is provided by someone pulling on the rope. Find the acceleration of the remaining block. c.) Explain why the aswers in a and b are different.



a= m2-m1 / m2+m1 * g



For part a, I found the following. a= 908-412 / 908+412 * 9.80 = 3.68m/s. Is this correct. For part b, how would the answer be different from part a if you are applying the same amount of force and pulling it in the same direction. I'm confused.

 

[AlephZero]

The question says "the heavier block was removed". So the new equation of motion only has one mass, not two.

In the first part, the tension in the rope was not 908N. If you draw a free body diagram for the heavy mass, you have its weight of 908N, the tension T in the opposite direction, and the resultant force of (908-T) causes the accleration of 3.68 m/s^2.

 

[m2003]

Hello,

Thank you for reaching out for help with this Newton's Law pulley problem. I am happy to assist you in understanding the concepts and arriving at the correct answers.

First, let's review the basics of Newton's Laws. The first law states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be represented by the equation F=ma, where F is the net force, m is the mass, and a is the acceleration.

In this problem, we have two blocks connected by a rope and pulleys. The first block has a weight of 412N and the second block has a weight of 908N. The rope and pulleys are massless and there is no friction, which means that the forces acting on the system are only the weight of the blocks and the tension in the rope.

a) To find the acceleration of the lighter block, we can use the equation F=ma and rearrange it to solve for acceleration (a=F/m). The net force acting on the lighter block is the difference between the weight of the heavier block (908N) and the weight of the lighter block (412N), so we have F=908N-412N=496N. The mass of the lighter block is also given as m=412N/9.80m/s^2=42.24kg. Plugging these values into the equation, we get a=496N/42.24kg=11.73m/s^2. So, the acceleration of the lighter block is 11.73m/s^2.

b) In part b, we are removing the heavier block and applying a downward force of 908N by pulling on the rope. In this case, the net force acting on the remaining block is the tension in the rope (908N) minus the weight of the block (412N), which gives us a net force of F=908N-412N=496N. The mass of the remaining block is still 42.24kg. Plugging these values into the equation F=ma, we get a=496N/42.24kg=11.73m/s^2. So, the acceleration of the remaining block is also