Understanding Variable Force and Kinetic Energy: Exploring Halliday's Answers

AI Thread Summary
The discussion revolves around solving a physics problem related to variable force and kinetic energy. The initial kinetic energy of a 2.0 kg body moving along the x-axis is 4.0 m/s, and the participants calculate the kinetic energy at various points based on the work done by the force. The correct answers are 12 J for part (a), 4.0 m for part (b), and 18 J for the maximum kinetic energy attained. Participants express confusion over the calculations, particularly regarding the work-energy principle and the interpretation of negative work. The importance of correctly identifying the initial kinetic energy is emphasized in resolving the problem.
mbrmbrg
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Behold Halliday's Question and Halliday's answers:

The only force acting on a 2.0 kg body as it moves along the x-axis varies as shown in Figure 7-41 (see attatchment). The velocity of the body at x = 0 is 4.0 m/s.

(a) What is the kinetic energy of the body at x = 3.0 m?
12 J
(b) At what value of x will the body have a kinetic energy of 8.0 J?
4.0 m
(c) What is the maximum kinetic energy attained by the body between x = 0 and x = 5.0 m?
18 J

I am stumped. I've only tried to figure out (a) and (b), but my method does not give me correct answers. My train of thought was that Delta KE=W=area under curve. So for part (a), I said that area under graph is (triangle-triangle-rectangle) = -4 Nm. Well, gee.
Maybe since KE is always positive, take the absolute value of various areas! I think that is very poor reasoning, because Delta KE and W can both be negative. And even when I do that, I get 8 Nm, which is still wrong.
Similar reasoning applied to (b) also gives the incorrect answer, which is comforting.

A hint from the wise gurus perhaps?
 

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mbrmbrg said:
My train of thought was that Delta KE=W=area under curve. So for part (a), I said that area under graph is (triangle-triangle-rectangle) = -4 Nm. Well, gee.

Correct.

The change of KE is equal to the work done on the body. If the work is negative, the kinetic energy will decrease. Final KE - initial KE = W. You know the initial KE, (16 J) you know W,(-4 J) what is the problem?

ehild
 
Problem is that I thought initial Kinetic Energy was 4J. Wups!
Thank you.
 
I'm doing the same problem. Does anyone know the answers?
I got 12J for (a) but I'm lost on (b) and (c)
 
Edit: this is what I got after some thought...

(a) 12 J
(b) x=4
(c) 28 J

anyone agree?
 
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