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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A 106g bullet is fired from a rifle having a barrel 0.271m long.Assuming the origin is placed where the bullet begins to move,the force exerted on the bullet by the expanding gas is F=a+bx-cx$^2$,where a=13500N, b=8170N/m,c=24000N/m$^2$,with x in meters.Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

$\begin{array}{1 1}3799.29J\\3699.29J\\1799.29J\\2699.29J\end{array} $

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Answer : 3799.29J
The work is found by integrating the force over the distance.
$W=\int\limits_{x_i}^{x_f} F.ds$
For the force in this problem we have,
$W=\int\limits_0^l (a+bx-cx^2)dx$
$\;\;\;\;=ax+\large\frac{1}{2}$$bx^2-\large\frac{1}{3}$$cx^3\big|_0^l$
$\;\;\;\;=(13500N)(0.271m)+\large\frac{1}{2}$$(8170N/m)(0.271m)^2-\large\frac{1}{3}$$(24000N/m^2)(0.271m)^3$
$\;\;\;\;=3799.29J$
answered Sep 1, 2014 by sreemathi.v
 

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