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

Find the work done by F as it moves from a to b.

$\begin{array}{1 1}625J\\224J\\650J\\525J\end{array} $

1 Answer

Answer : 625J
$W=\int\limits_a^b\overrightarrow{F}. \overrightarrow{ds}$
$\;\;\;\;=\int\limits_0^{10\cos 30} 30xdx+\int\limits_0^{10 \sin 30} 40y \cos 180 dy$
$\;\;\;\;=\large\frac{30x^2}{2}\big]_0^{8.66}-\large\frac{40y^2}{2}\big]_0^5$
$\;\;\;\;=1125-500$
$\;\;\;=625J$

 

answered Sep 1, 2014 by sreemathi.v
edited Sep 2, 2014 by balaji.thirumalai
 

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