Work of an Object moving down Inclined Plane

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SUMMARY

The discussion focuses on calculating the work done by a suspended mass in a pulley system attached to a block on an inclined plane. The setup involves a 208g weight and a block weighing 474g on a 30-degree incline, with the weight moving down 24.7cm at a constant velocity. The work done by gravity is analyzed using the formula w=Fdcos(θ), emphasizing that the work is negative due to the opposing directions of force and motion. The conversation highlights the importance of expressing calculations symbolically before substituting numerical values to ensure clarity and accuracy.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of work-energy principles (w=Fdcos(θ))
  • Familiarity with gravitational potential energy calculations
  • Basic algebra skills for symbolic manipulation
NEXT STEPS
  • Study the concept of conservation of energy in mechanical systems
  • Learn how to calculate work done by gravity in inclined planes
  • Explore the implications of constant velocity on net force and work
  • Practice solving problems involving pulley systems and inclined planes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of work and energy concepts in real-world applications.

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Homework Statement


Experiment set up.
A 208g weight is attached via a pulley system to a block on an inclined plane. What is the work done by the suspended mass as the car is lowering at a constant velocity? And work done by gravity?

Distance weight moves down - 24.7cm
Incline is 30 degrees

Homework Equations


[itex]F=ma[/itex]
[itex]w=Fdcos(\theta)[/itex]


The Attempt at a Solution


So, in this case, work would be negative, right? Because the direction of the force from the suspended mass is going UP the incline, and the direction of moment is DOWN the incline?

The block is 474g

I'm not sure where to start. If I use the basic work formula and do
[itex]\(-208*980)*24.7[/itex]?
 
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The thing losing energy is doing the work.
The description of the experiment is incomplete - we are told about the suspended mass and then some car is introduced out of nowhere... but it looks like you are expected to use conservation of energy. Your descriptions of what you have tried are also incomplete so it is not clear what you have done.

Try expressing your working symbolically - do all the algebra before you put numbers in.
 
Simon Bridge said:
The thing losing energy is doing the work.
The description of the experiment is incomplete - we are told about the suspended mass and then some car is introduced out of nowhere... but it looks like you are expected to use conservation of energy. Your descriptions of what you have tried are also incomplete so it is not clear what you have done.

Try expressing your working symbolically - do all the algebra before you put numbers in.

Sorry, I wrote it really quickly.

The setup was an inclined plane with a pulley system which was set off the leg of the triangle. The pulley system was attached to the car. We had to set a weight on the pulley so that the car would go down the plane with a constant velocity.

I'm really not sure what to do. The car is moving down the car, therefore losing potential energy. If work can be defined as the difference in U.

So, if then ΔU= (474*980*12.5)-(474-980*0)= 5,806,500 erg?
 

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