Work, Energy, or Work problem

In summary, the problem involves an Atwood's machine with two masses initially at rest at the same height. After being released, the large mass, m2, falls through a height h and hits the floor while the small mass, m1, rises through the same height. The question asks for the speed of m2 just before it lands, given h = 1.2m, m1 = 3.7kg and m2 = 4.1kg. To solve this, the conservation of energy theorem is used, setting the reference level at the dotted line where v=0. The initial total potential and kinetic energies are both zero. In the final configuration, m1 has climbed up distance h, giving it a
  • #1
LastXdeth
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Homework Statement



WorkEnergyPowerQuiz2picture.png

The two masses in the Atwood’s machine shown in Figure 8–23 are initially at rest at the same height. After they are released, the large mass, m2 falls through a height h and hits the floor, and the small mass, m1rises through a height h. Find the speed of the masses just before m2 lands if h = 1.2m, m1 = 3.7kg and m2 = 4.1kg


Homework Equations



E=Eo
Ug=mgh
K=(1/2)mv^2


The Attempt at a Solution



m1*g*y +m2*g*y = (1/2)*m1*v^2 + (1/2)*m2*v^2

I factored outed g*y and (1/2)*m*v^2
g*y*(m1 + m2) = (1/2)*(v^2)*(m1 + m2)

I canceled out (m1 + m2)
g*y = (1/2)*v^2

g*y = (1/2)*v^2

2*g*y = v^2

√(2*g*y) = v

√(2*9.8*1.2) = v

This is apparently wrong:
4.85m/s ≈ v
 
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  • #2
first define the reference level for the potential energy... what level have you chosen ?
 
  • #3
I would like to set the reference level at where the dotted line is (where v=0).
 
  • #4
ok, so that means your initial total potential energy is zero. Also initial total kinetic energy
is zero. So total initial energy is zero. What about the final configuration ? The left block has climbed up distance h , so its potential energy is [itex]m_1gh[/itex] and the second
block is gone below the reference level. so its potential energy would be
[itex]-m_2gh[/itex]. What about their kinetic energies ? Find that expression and use the conservation of energy theorem, which implies that the total initial energy must be equal to the total final energy...
 
  • #5



As a scientist, it is important to approach problems like these with a clear understanding of the concepts involved. In this case, the problem involves the concepts of work and energy. Work is defined as the force applied over a distance, while energy is the ability to do work. In this problem, the two masses are initially at rest, meaning they have no kinetic energy. As they fall and rise, work is being done on them by the force of gravity, which is converting potential energy into kinetic energy. This can be represented by the equation:

W = ΔE = ΔK + ΔU

Where W is the work done, ΔE is the change in energy, ΔK is the change in kinetic energy, and ΔU is the change in potential energy.

We can also use the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the initial potential energy of the system is equal to the final kinetic energy of the masses before m2 hits the ground. This can be represented by the equation:

Ei = Ef

m1gh = (1/2)m1v1^2 + (1/2)m2v2^2

Solving for v2, we get:

v2 = √(2gh)

Plugging in the given values, we get:

v2 = √(2*9.8*1.2) = 4.85 m/s

This is the speed of the masses just before m2 hits the ground. It is important to note that this is the final speed of the system, as all of the potential energy has been converted into kinetic energy.
 

What is work?

Work is defined as the product of force and displacement in the direction of the force. In simpler terms, work is the energy transferred to or from an object by means of a force acting on the object. It is measured in joules (J).

What is energy?

Energy is the ability to do work. It can take various forms, such as kinetic energy (energy of motion), potential energy (stored energy), thermal energy (heat), and many others. Energy is measured in joules (J).

What is the difference between work and energy?

Work and energy are closely related concepts, but they are not the same. Work is the transfer of energy, while energy is the capacity to do work. Work is a scalar quantity, while energy is a vector quantity. Additionally, work is dependent on both force and displacement, while energy is dependent on the state and position of an object.

What is a work problem?

A work problem is a type of physics problem that involves calculating the amount of work done by a force on an object. These problems usually require knowledge of the force acting on the object, the displacement of the object, and the angle between the force and displacement vectors.

How do I solve a work problem?

To solve a work problem, you will need to use the formula W = Fd cosθ, where W is work, F is force, d is displacement, and θ is the angle between the force and displacement vectors. Make sure to convert all units to the appropriate SI units (joules for work, newtons for force, and meters for displacement) before plugging in the values. Then, simply solve for W to find the work done by the force on the object.

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