Sliding down a ramp against friction force

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Homework Help Overview

The problem involves a body sliding down a ramp while experiencing a friction force. The original poster seeks to determine the change in kinetic energy after the body travels a specified distance down the ramp.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy, frictional energy, and kinetic energy. There are attempts to set up equations to express the change in kinetic energy based on energy conservation principles. Some participants question how to formulate the equation for this change.

Discussion Status

The discussion is ongoing, with participants providing insights into the energy relationships involved. Some guidance has been offered regarding the approach to calculate the change in kinetic energy, but no consensus has been reached on the specific formulation of the equations.

Contextual Notes

There is an emphasis on the work done by friction and its impact on the change in kinetic energy. The original poster expresses confusion about setting up the necessary equations, indicating a potential gap in understanding the concepts involved.

Vader07
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Homework Statement



At the top of the ramp the body is stopped and then released. It slides down the ramp against a friction force of f= 10 Newtons. Find the body's change in kinetic energy in Joules after traveling a distance s= 3 m down the ramp.


Homework Equations




There are three types of energy:

1) potential energy: Epot : the body is on top of the ramp (height h) at rest. The energy is Epot = mgh
2) frictional energy: Efric = cmg*cos alpha *s (= frictional force multiplied by the distance s that the body had gone down the ramp at a certain point.)
3) kinetic energy: Ekin = 1/2 mgv^2 : let's you calculate the velocity of the body.

The three energies are related in the following way:
Epot = Efric + Ekin, -->
Ekin = Epot - Efric .


The Attempt at a Solution



KE = 1/2mgv^2
KE = 1/2(1)(9.8)^2
= 1/2(96.04)
= 49.04

F1 = (1)(9.8)sin(alpha)
 
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the change in kinetic energy is simply going to be your change in potential energy minus the energy lost to friction(Or the work done by friction). With emphasis on work.
 
dacruick


thanks for the reply. could you help me set the equation for this change in kinetic energy. I am sort of lost on this.
 
just take the kinetic energy at the start and the kinetic energy at the end, and then find the difference. do you know the equation for kinetic energy?
 

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