# Algebra of pulley-mass energy equation

• PacFan01
In summary: So, what you are saying is that the block has the same speed as the string, even though it is farther away?
PacFan01

## Homework Statement

[/B]
A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disc of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.30 m off the ground. What speed does this block have when it hits the ground?

## Homework Equations

ΔE = 0
Ei = Ef (closed system)
E = K + UG + Eth + Es
UiG = Ktransf + Krotf
Ktrans = ½mv2
Krot = ½Iω2
ISolid Cylinder = ½mr2
ω = v/r

## The Attempt at a Solution

[/B]
First, I step up the energy equation and substitute in my energy expressions. I've used the subscript "1" to represent the 0.20-kg mass, the subscript "2" to represent the 0.25-kg mass, and the subscript "p" to represent the pulley.

U1iG + U2iG = U1fG + U2fG + K1f + K2f + Kpf

m1gh1i + m2gh2i = m1gh1f + m2gh2f + ½m1v12 + ½m2v22 + ½Iω2

I know what the next step is, but I'm not sure how to get there algebraically.

I should end up with:

(m1 - m2)gΔh = ½(m1 + m2)vf2 + ¼mpvf2

I understand that v1, v2, and vp are equal, so no problem there. I see that how we substitute ½mr2 for I, and v/r for ω. I don't understand how we end up with (m1 - m2)gΔh. I understand that we pull the common variable g out, and I can kind of understand how the h-values become Δh. However, it looks like this when I set it up:

m1gh1i + m2gh2i - m1gh1f - m2gh2f = ½m1v12 + ½m2v22 + ½Iω2

Which becomes:

g(m1h1i + m2h2i - m1h1f - m2h2f) = ½m1v12 + ½m2v22 + ½Iω2

Any help is greatly appreciated.

The length of the rope does not change. If the 0.25 kg block descends by 0.3 m, the other block rises by 0.3 m.
(h1f-h1i= h2i-h2f)

I initially responded with another question, but I think I understand now.

## What is the Algebra of Pulley-Mass Energy Equation?

The Algebra of Pulley-Mass Energy Equation is a mathematical equation that relates the mass and acceleration of an object to the tension and distance of a pulley system. It is used to calculate the mechanical energy involved in the motion of objects connected by pulleys.

## How is the Algebra of Pulley-Mass Energy Equation derived?

The equation is derived from the principles of conservation of energy and Newton's laws of motion. It takes into account the work done by the weight of the object, the tension in the rope, and the distance the rope is pulled.

## What are the components of the Algebra of Pulley-Mass Energy Equation?

The equation consists of the mass of the object (m), the acceleration due to gravity (g), the tension in the rope (T), and the distance the rope is pulled (d). It can be written as E = mgh + Td, where E is the total mechanical energy.

## How is the Algebra of Pulley-Mass Energy Equation used in real-life situations?

The equation is commonly used in physics and engineering to analyze and design systems that involve pulleys. It can be used to determine the amount of force or work needed to move an object connected to a pulley system, as well as the speed and acceleration of the object.

## What are some common mistakes made when using the Algebra of Pulley-Mass Energy Equation?

Some common mistakes include not accounting for the weight of the pulley itself, not considering the direction of motion (up or down), and using incorrect units. It is important to carefully label and define all variables in the equation before solving for the desired quantity.

• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
21
Views
9K
• Introductory Physics Homework Help
Replies
66
Views
5K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
3K