Consider an object of mass $5kg$ on a plane. If a horizontal force of $10N$ acts on the object for a time interval of a second. Calculate the work done on the object by the force given that the force of kinetic friction on the object is $5N$.

The acceleration due to the net force is $1m/s^2$. So the distance moved in the direction of the force when the force was being applied is $0.5m$.

Here is what I am confused about. While calculating the work using the formula $$W=F.s$$ should we plug in the net force on the object that is the (Horizontal force exerted - the frictional force) or should we just plug in the horizontal force?

  • $\begingroup$ Please see our guide on writing good titles. $\endgroup$ – user10851 Apr 16 '16 at 18:31

You should plug in the horizontal force acting on the object if you want to know the work done by this force. Due to friction heating up the plane not all of this work is converted in the objects kinetic energy. PS: I assume you meant "while calculating the work" rather than "while calculating the force"..

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  • $\begingroup$ OK, that makes sense. Is there something called the "net" work done on an object? $\endgroup$ – The Cryptic Cat Apr 15 '16 at 4:12
  • $\begingroup$ It seems that often "net" work is defined as only the change in kinetic energy of the object (hence ignoring frictional losses) which would indeed be obtained by subtracting the frictional force from the horizontal force exerted on it. $\endgroup$ – Jan Bos Apr 15 '16 at 4:30

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