How Is Work Calculated With Friction?

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Work is calculated by considering the net force acting on an object, which in this case involves an applied force of 25N and a frictional force of 10N. The net force is thus 15N, but since the chair moves at a constant speed, the effective force doing work against friction is 10N. To find the work done, the equation Work = Force * Displacement is applied, resulting in Work = 10N * 2.5m, which equals 25J. The discussion clarifies that the force of friction is the key factor in calculating work when an object moves at constant speed. Understanding the relationship between applied force and friction is essential for solving similar physics problems.
rice1am
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Hey guys,
I got a test tommorow regarding Work, Efficiency, Conservation of Mass, and all those other things. And I needed help with this question regarding work with friction. Oooh, I wish I was good in physics :( If anyone knows the answer, please post and tell me how to do it properly. Many thanks!

Homework Statement


A toddler pushes a chair at a constant speed with a force of 25.0N (Applied Force = 25N to the right) for a distance of 2.5m, but the chair is being pushed across a smooth floor against a force of friction (resistance of 10.0N) (Force of friction = 10N to the left). How much work is being done?


Homework Equations


Work=AppliedForce*displacement

Force of gravity=mg

Net force=ma


The Attempt at a Solution


Well, I figured that I would subtract 10N from 25N and then use the result (15) and calculate it by the displacement, which is 2.5m. But when I checked my answer at the back of my physics textbook, my answer was incorrect. Now, I'm pretty much stuck? Would the force of friction be of the same value as the applied force?
 
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What they give you is the force against which you are doing work.

It's 10N.

That's what you use to calculate the Work.

Now the chair is not being accelerated so evidently the magnitude of the force the toddler is supplying is greater than needed if it was all in the direction of motion. This apparently means then that the force is at an angle and the horizontal component of it is 10N.
 
oh, I see.

So, would the equation be as follows?

Work= Applied Force * Displacement
= 10N * 2.5m
= 25J
 
rice1am said:
oh, I see.

So, would the equation be as follows?

Work= Applied Force * Displacement
= 10N * 2.5m
= 25J

That looks correct.
 
Thanks! :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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