Calculating Work Using a Force vs Distance Graph

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The discussion focuses on calculating work using a force versus distance graph, with the formula W = (Force component * displacement) applied to determine work done. The first segment from 5m to 10m has a constant force of 30N over a 5m displacement, resulting in 150J of work. The second segment from 10m to 15m shows a varying force, starting at 30N and dropping to approximately 18N, calculated using the area of a triangle, yielding 120J. The total work calculated is 270J, which is confirmed as correct by another participant. The conversation emphasizes the importance of using area calculations for varying forces on a graph.
Le_Anthony
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Homework Statement


queston 9.png

Homework Equations


W=(Force component*displacement)

The Attempt at a Solution


This problem uses a graph to find the Work. So I used area equations to find the Work. From 5m to 10m the Force is constant 30N. Which is applied over a displacement of 5m. So the work done from 5m to 10m is 150J. Next, from 10m to 15m the work varries. Its initially 30N then it drops down to roughly 18N. So using a triangle, .5*(30N+18N)*(5m)=120J

Answer is 270J??
 
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Hi Le:

Your answer looks OK to me.

Regards,
Buzz
 
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Buzz Bloom said:
Hi Le:

Your answer looks OK to me.

Regards,
Buzz
Thanks Buzz!
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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