Energy (Potential, kinetic ) So stuck

[zomething]

Energy (Potential, kinetic...) So stuck :(

Homework Statement

Two objects, m1 = 5.50 kg and m2 = 3.00 kg, are connected by a light string passing over a light frictionless pulley as shown in the figure below. The object of mass 5.50 kg is released from rest, h = 4.50 m above the ground.

(a) Using the isolated system model, determine the speed of the 3.00 kg object just as the 5.50 kg object hits the ground. (I got 5.10 m/s)

(b) Find the maximum height to which the 3.00 kg object rises.

Homework Equations

K1+U1=K2+U2

The Attempt at a Solution

Part a was not a problem. It is part b that I'm stuck on. Common sense tells me that the maximum height that it will reach is equal to the height from which m2 is dropped. Apparently, my logic is faulty. So I tried the K1+U1=K2+U2 approach and am getting the wrong answer. I tried assuming that m2 had no potential energy when it reaches the bottom (and m1 would reach its maximum height), but wouldn't m2 also have no kinetic energy at the bottom, and m1 no kinetic energy at the top? And besides that, wouldn't their velocities be equal to zero, leaving me only with U1? I've literally been trying to figure this out for the last hour, what am I not seeing? Any help greatly appreciated!

 

[tiny-tim]

Welcome to PF!

Hi zomething! Welcome to PF!

It is part b that I'm stuck on. Common sense tells me that the maximum height that it will reach is equal to the height from which m2 is dropped.

he he

when m₂ hits the ground, the rope will go slack (ie no tension), so m₁ will just be a projectile, launched upward …

find the speed of m₁, and then use conservation of energy.

(alternatively, regard m₂ as continuing to have the same KE as it had just before it hit the ground)

 

[kerimek]

It seems like you are on the right track with using the conservation of energy equation, but there may be some confusion with the initial and final states of the system. Let's break it down step by step:

1. Initially, the system has a potential energy of m1gh where h = 4.50 m.
2. As m1 falls, it loses potential energy and gains kinetic energy. At the same time, m2 rises and gains potential energy.
3. At the moment m1 hits the ground, it has lost all of its potential energy and gained kinetic energy equal to 1/2m1v^2. At this point, m2 has also reached its maximum height and has no kinetic energy.
4. Using the conservation of energy equation, we can set the initial and final states of the system equal to each other:
m1gh = 1/2m1v^2 + m2gh
5. Rearranging for v, we get v = √(2gh(1-m2/m1)).
6. Plugging in the values given, we get v = 5.10 m/s.
7. Now, to find the maximum height reached by m2, we can use the equation for potential energy, mgh, and set it equal to the initial potential energy of the system (m1gh) minus the final potential energy of m2 (m2gh):
mgh = m1gh - m2gh
8. Rearranging for h, we get h = (m1/m2-1)h.
9. Plugging in the values given, we get h = 3.00 m, which is the maximum height reached by m2.

I hope this helps clarify the problem and your solution! It's important to keep track of the initial and final states of the system and how energy is transferred between objects. Keep up the good work!