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

A particle of mass 0.5 kg moves on the positive $x$-axis under the action of a variable force $\large\frac{40}{x^2}$ newtons,directed away from origin..The particle passes through a point 2 metres from origin,with velocity $8ms^{-1}$ away from origin. Its experiences a constant resistance force of 6 newtons. Find the speed of the particle when it is 5 metres from origin.

$\begin{array}{1 1}\sqrt{40}ms^{-1}\\\sqrt{30}ms^{-1}\\\sqrt{15}ms^{-1}\\\sqrt{13}ms^{-1}\end{array} $

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Answer : $\sqrt{40}ms^{-1}$
The work done by the resistance is $6\times 3 =18J\qquad$ Decreases K.E so negative
The work done by the force is $\int\limits_{2^5} \large\frac{40}{x^2}$$dx$
$\Rightarrow \big[\large\frac{-40}{x}\big]_2^5$$=12J\qquad$ Increases K.E so positive
Final K.E = Initial K.E-work done by resistance + work done by force
$\Rightarrow \large\frac{1}{2}$$\times 0.5V^2=\large\frac{1}{2}$$\times 0.5\times 8^2-18+12=10J$
$\Rightarrow V=\sqrt{40}ms^{-1}$
answered Sep 1, 2014 by sreemathi.v

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