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

A force $F_x$ acts on a particle.The force is related to the position of the particle by the formula $F_x=Cx^3$,where $C$ is a constant.Find the work done by this force on the particle when the particle moves from $x=1.5\;m$ to $x=3\;m$.

$\begin{array}{1 1}19C\;J\\29C\;J\\39C\;J\\10C\;J\end{array} $

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1 Answer

Answer : 19C J
$W=\int Fdx$
$W=C\int\limits_{1.5}^3x^3dx$
$\;\;\;\;=\large\frac{C}{4}$$(3^4-1.5^4)$
$\;\;\;\;=19C\;J$
answered Sep 1, 2014 by sreemathi.v
edited Sep 2, 2014 by sreemathi.v
 

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