Confusion about work done by friction as negative or positive

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Discussion Overview

The discussion revolves around the interpretation of work done by friction in the context of a physics problem involving two blocks, one of which is subject to kinetic friction. Participants explore the implications of using conservation of energy and the work-energy principle to analyze the system, particularly focusing on the signs associated with work done by friction and potential energy changes.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an equation using conservation of energy and work done by friction, questioning why the work done by friction appears positive in their calculations despite being conceptually negative due to opposing directions of force and displacement.
  • Another participant challenges the sign of the work done by friction, suggesting a different approach to the equation that incorporates non-conservative forces, leading to a different interpretation of energy changes.
  • A third participant proposes an alternative method that aligns with the previous equations but emphasizes the relationship between work done by friction and changes in kinetic energy.
  • Further contributions clarify that removing friction results in a larger kinetic energy, reinforcing the idea that friction does negative work and affects the overall energy balance in the system.
  • One participant expresses understanding after reviewing the equations and discussions presented.

Areas of Agreement / Disagreement

Participants express differing views on the sign convention for work done by friction and its implications for energy calculations. There is no consensus on the correct interpretation, as multiple perspectives on the equations and their outcomes are presented.

Contextual Notes

Participants rely on specific assumptions about the system, such as the values for mass and gravitational acceleration, which may influence their calculations. The discussion highlights the complexity of applying conservation principles in the presence of non-conservative forces like friction.

Who May Find This Useful

This discussion may be useful for students and educators in physics, particularly those grappling with concepts of work, energy, and the effects of friction in mechanical systems.

kjamha
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https://lh3.googleusercontent.com/0GOyZeQMdpLgYxYwwP4FMabe-Mk7QpADPC4ZK4PVpBGKvamcqjopOf3M0evdhE8PHZ3iTQ=s138 m1 is 4 kg and m2 is 8 kg. The kinetic friction coefficient on the table is 0.3. m1 is held in place. When m1 is released, m2 accelerates 1.2 m to the floor. Use conservation of energy/ and or Work KE principle to find the speed of the two blocks at the instant m2 hits the floor.

I used the following equation to solve the problem:

PE = KE1 + KE2 + work(friction) with g = 10 m/s2

8x10x1.2 = 1/2 (4) (v2) + 1/2 (8) (v2) + Nx.3x1.2 and then solved for v.

I got the right answer but I see work done by friction as negative (the force and displacement are in opposite directions). If I use a negative number here, I get the wrong answer. Also, regarding conservation of energy, I see a change in PE as negative energy (decrease in PE), the friction as negative work and the change in KE for m1 and m2 as a gain in energy. Shouldn't the change in PE + work done by friction = change in KE of m1 + change in KE of m2? Why is work done by friction positive in my original equation?
 
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You've got the sign of Wf wrong. How did you arive at this equation? Because when non conservative forces are present, ΔΕ=W, where W is the work of the non conservative forces. ΔΕ=K1+K2-P0, so P0=K1+K2-Wf.
 
Another method giving the same result is this:
ΔΚ=Wf+Wg
Wg=-ΔP=P0
P0=K1+K2-Wf
 
Your equation makes sense to me. Plugging in numbers for P0=K1+K2-Wf (using g = 10m/s2):
96 = 4v2 + 2v2 - 0.03x40x1.2
96 + 14.4 = 6v2
v = 4.29 m/s
If friction is eliminated, then 96 = 4v2 + 2v2 and v = 4 m/s. This is slower than 4.29. Adding friction should slow down the speed. That is why added Wf, which gives me the correct answer. I like your equations, but why is it giving me the wrong answer?
 
P0+Wf=K (where K=K1+K2), and since friction does a negative work, if you remove it, K is larger.
 
I see it. Thanks!
 
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