Calculating the displacement using the work-energy principle

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SUMMARY

The discussion focuses on calculating the displacement of an object traveling up an inclined surface at an initial speed of 10 m/s and an angle of 10°, considering a coefficient of friction of 0.15. The work-energy principle is applied, utilizing equations such as Ei = Ef and W = Fd. The user successfully determines the displacement to be approximately 32.6 meters, while also addressing the need to incorporate the kinetic friction coefficient in their calculations. The discussion concludes with a verification of the approach to find the final velocity upon descent.

PREREQUISITES
  • Understanding of the work-energy principle
  • Familiarity with free body diagrams (FBD)
  • Knowledge of basic kinematics and dynamics
  • Proficiency in solving equations involving friction
NEXT STEPS
  • Study the implications of kinetic friction in energy loss calculations
  • Learn how to derive equations for motion on inclined planes
  • Explore advanced applications of the work-energy principle in different scenarios
  • Investigate the effects of varying angles and coefficients of friction on displacement
USEFUL FOR

Students in physics, particularly those studying mechanics, as well as educators and anyone interested in applying the work-energy principle to real-world problems involving inclined surfaces and friction.

brycenrg
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Homework Statement


An object is traveling at 10m/s. along an inclined surface of an angle 10°. How far does it go up the bank?
How fast will it be going when it travels back down?
Coefficient of friction is .15

Homework Equations


Ei = Ef
Ef - Ei = Eloss
[/B]
W = Fd

The Attempt at a Solution


I don't really know where to start. I made a FBD.
∑Fx = gsinθ+ukgcosθ = ma
There are two unknowns. I don't know the distance. and I don't know how much energy is lost in the displacement.

If there was no friction I could solve it using Ei = Ef but with friction I don't know where tostart. Can anyone tell me the first step?[/B]
 
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I tried again. 1/2mv^2 = ukmg cos∅d+mgh
h = d sin∅
(v^2)/(2g(cos∅+sin∅)) = d =32.6 m
That is the displacement. I think.
use h = dsin∅ to find initial H
use mgh = ukmgcos∅d+1/2mv^2 to find v
Is that correct?
 
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brycenrg said:
I tried again. 1/2mv^2 = ukmg cos∅d+mgh
h = d sin∅

That's a good start.

It looks like you left μk out of the following equation.
(v^2)/(2g(cos∅+sin∅)) = d =32.6 m
That is the displacement. I think.
use h = dsin∅ to find initial H
]use mgh = ukmgcos∅d+1/2mv^2 to find v
Is that correct?
 
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