We know ${\rm work} = \int F \cdot dl$ , where $dl$ is the displacement of point of application of force.

Suppose I apply a instant force of $100~\rm N$ on a body and then remove the force; after that the body moves a distance $d$. When the body is moving there is no external force (other than friction, $mg$, normal reaction force). So what is the work done by $100~\rm N$ force?

Is is zero? Because when the body moved distance $d$, the $100~\rm N$ force was not there.

  • $\begingroup$ If we draw a free body diagram while the body is moving, there will be only forces like mg, normal,air drag and no external force, so where is the work done by that force??I am just confused a lot $\endgroup$ – pik selvan Jun 9 '18 at 6:38
  • $\begingroup$ In the formula $W=\vec{F} . \vec{l} $ , $l$ is zero , cause it is the distance traversed as long as $F$ acts ... $\endgroup$ – Nehal Samee Jun 9 '18 at 8:28
  • $\begingroup$ Thanks , thats a perfect reply.dl should be calculated as long as F acts. $\endgroup$ – pik selvan Jun 9 '18 at 8:49

$d=0$ in your example, because the work is zero and so the kinetic energy of the body is zero. A finite force applied for an instant doesn't do any work.

  • $\begingroup$ I explain my question further.Suppose a 100 N force is applied on a body and due to this 100 N force, body moves 10 m. After moving 10 m, the 100 N force is removed, and body moves 5 m further before coming to rest. what is the work done by the 100 N force :;; W=100*10=1000 J or W= 100 *(10+5)=1500 J. please guide Mr Chris $\endgroup$ – pik selvan Jun 9 '18 at 8:57
  • $\begingroup$ It's $1000 J $ work ... $\endgroup$ – Nehal Samee Jun 9 '18 at 9:37

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