How Does the Conservation of Energy Explain the Motion of Objects Over a Pulley?

[Leesh09]

Homework Statement

Two objects are connected by a light string passing over a light frictionless pulley as shown
in the figure. The object of mass 5.00-kg is released from rest. Using the principle of
conservation of energy, (a) determine the speed of the 3.00-kg object just as the 5.00-kg
object hits the ground. (b) Find the maximum height to which the 3.00-kg object rises.

Homework Equations

The Attempt at a Solution

I believe I am all set with part a. My question is on part B. Is it correct to say 1/2*m*v² = m*g*h and then use 9.8 as v also and solve for h?

 

[kuruman]

The statement of the problem is not very informative without a figure. Can you post one or at least describe what the figure shows?

 

[kerimek]

I would like to clarify that the principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the potential energy of the 5.00-kg object is converted to kinetic energy as it falls, and then transferred to the 3.00-kg object as it rises. Therefore, the total energy (potential + kinetic) of the system remains constant throughout the motion.

For part (a), you can use the conservation of energy equation: m1gh1 = m2gh2 + 1/2m2v2^2. Since the 5.00-kg object is released from rest, its initial velocity is 0 and its initial potential energy is m1gh1 = 0. The final potential energy of the 3.00-kg object is also 0 at its maximum height, leaving us with the equation: 1/2m2v2^2 = m1gh1. Solving for v2, we get v2 = √(2m1gh1/m2). Plugging in the values, we get v2 = √(2*5.00*9.8/3.00) = 8.16 m/s.

For part (b), you can use the same equation but this time, you have to solve for h2. The equation becomes: m1gh1 = m2gh2 + 1/2m2v2^2. Solving for h2, we get h2 = (m1gh1 - 1/2m2v2^2)/m2. Plugging in the values, we get h2 = (5.00*9.8 - 1/2*3.00*8.16^2)/3.00 = 6.56 m.

So, the maximum height to which the 3.00-kg object rises is 6.56 m. I hope this helps!