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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A 106g bullet is fired from a rifle having a barrel 0.441m 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.With what velocity does the bullet exit the barrel?

$\begin{array}{1 1}6061.83J\\6051.83J\\6000J\\2699.29J\end{array} $

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1 Answer

Answer : 6061.83J
The work is found by integrating the force over the distance
$W=\int\limits_{x_i}^{x_f} F.ds$
For the barrel of length $l_n=0.441m$,we need only change the limits of integration,
$W=\int\limits_0^{l_n} (a+bx-cx^2)dx$
$\;\;\;\;=ax+\large\frac{1}{2}$$bx^2-\large\frac{1}{3}$$cx^3\big|_0^{l_n}$
$\;\;\;\;=(13500N)(0.441m)+\large\frac{1}{2}$$(8170N/m)(0.441m)^2-\large\frac{1}{3}$$(24000N/m^2)(0.441m)^3$
$\;\;\;\;=6061.83J$
answered Sep 1, 2014 by sreemathi.v
 

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