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  • A particle experiences a variable force <img alt="\vec{F}\mathrm{= \left ( 4x\, \hat{i}+3y^{2}\, \hat{j} \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cvec%7BF%7D%5Cmathrm%7B%3D

A particle experiences a variable force \vec{F}\mathrm{= \left ( 4x\, \hat{i}+3y^{2}\, \hat{j} \right )} in a horizontal \mathrm{x-y} plane. Assume distance in meters and force is newton. If the particle moves  from point \mathrm{\left ( 1,2 \right )} to point \mathrm{\left ( 2,3 \right )} in the \mathrm{x-y} plane, then Kinetic Energy changes by :

Option: 1

\mathrm{50.0\: J}


Option: 2

\mathrm{12.5\: J}


Option: 3

\mathrm{25.0\: J}


Option: 4

\mathrm{0\: J}


Answers (1)

best_answer

Force experienced by particle in a honizontal X-Y plane is,
\bar{F}=4 x\left ( \hat{i} \right )+3 y^{2} \left ( \hat{j} \right )
Comparing with,
\bar{F}= F_{x}\hat{i}+F_{y}\hat{j}
\mathrm{F_{x}= 4x, \quad F_{y}= 3y^{2}}

\mathrm{dW}= \bar{F}\cdot d\bar{s}
         =\left ( F_{x}\hat{i}+F_{y}\hat{j} \right )\cdot \left ( d_{x}\, \hat{i}+d_{y}\, \hat{j} \right )
\mathrm{dW=F_{x}\, dx+F_{y}\, dy}

\mathrm{W= \int dW= \int_{x=1}^{x= 2}4x\, dx+\int_{y=2}^{y= 3}3y^{2}\, dy= \Delta KE}

(from Work-energy theorem)
\mathrm{W =4\left[\frac{x^{2}}{2}\right]_{1}^{2}+3\left[\frac{y^{3}}{3}\right]_{2}^{3}}
      \mathrm{ =4\left [ 2-\frac{1}{2} \right ]+3\left [ 9-\frac{8}{3} \right ]}
\mathrm{ W=6+19= 25J= \Delta KE}

The correct option is (3)

 

 

Posted by

shivangi.shekhar

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