Mechanical Energy Problem/non-conservative forces

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SUMMARY

The discussion centers on calculating the constant friction force (F) acting on a body of mass "m" as it slides down a surface and then travels horizontally. The user correctly applies the work-energy principle, stating that the change in kinetic energy equals the work done by friction. The derived formula for friction force is F = (10mh)/5 = 2mh, which is accurate given the parameters of gravitational potential energy and the distance traveled. Feedback from other users highlights the importance of clarity in variable definitions and sign conventions.

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fuvest
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Homework Statement


A body of mass "m" is let go from on top of a surface A, where it slides down to B(without friction). From that point on, it displaces itself on an horizontal surface 5 meters away from B, where it stops at C.
Being "m" a mass in kg
"h" in meters and g = 10 m/s^2
The value, in Newtons, of the constant friction force F when the body dislocates itself is:

Homework Equations


Ei = Ef
W = F.d

The Attempt at a Solution


My attempt was to set the change in kinetic energy equal to the total work. So therefore:
1/2mv^2 - mgh(converted from gravitational potential energy) = Tf
at the very end, velocity will be zero because it stops so:
-mgh = Tf
replacing the work done by friction:
-mgh = -F.5
.'. F = (10mh)/5 = 2mh

Is this approach correct? Would you guys suggest another one? Is another possible?
Thanks
 
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fuvest said:
Is this approach correct?
Looks fine. Not sure about some of the signs, but since you have not stated your sign convention they may be fine.

For what it's worth, it is very poor style to specify that a variable represents the numerical value of a physical quantity when expressed in certain units ("m being a mass in kg..."). It would have been better to statethe horizontal distance as another unknown, s say, and not mention units at all. The answer would then have been mgh/s.
 

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