Area under force vs displacement graph :D?

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SUMMARY

The discussion focuses on calculating the final kinetic energy (Kf) at a displacement of 4 meters using the work-energy principle. The user correctly computes the work done (W) as 4 Joules by finding the area under the force vs. displacement graph, applying the formula W = F.d. They then use the equation W = Kf - Ki, substituting the initial kinetic energy (Ki) of 2 Joules to arrive at a final kinetic energy (Kf) of 6 Joules. The calculation is confirmed as accurate by other forum members.

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DK_sam
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Homework Statement


Find K at 4m, besides the F.s graph they also give me Kinitial = 2J

Homework Equations


W=F.d and W=Kf- Ki

The Attempt at a Solution


The first thing I did was to find the area under the graph.
1/2(4)(2)=4J W=F.d so that's my work.
Then I replaced that 4J into the other eq W= Kf-Ki
then solved for Kf.
W+Ki=Kf
plugging in values 4j +(2J)=Kf
then 6J=Kf
Is that correct guys? Thanks for your help!
 

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DK_sam said:

Homework Statement


Find K at 4m, besides the F.s graph they also give me Kinitial = 2J

Homework Equations


W=F.d and W=Kf- Ki

The Attempt at a Solution


The first thing I did was to find the area under the graph.
1/2(4)(2)=4J W=F.d so that's my work.
Then I replaced that 4J into the other eq W= Kf-Ki
then solved for Kf.
W+Ki=Kf
plugging in values 4j +(2J)=Kf
then 6J=Kf
Is that correct guys? Thanks for your help!
looks good! Always put a space between the numerical value and the unit, 6 J not 6J.
 
Thank you good sir! I will next time! :D good night!
 

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