Question

# A block of mass $10kg$ is moving in x-direction with a constant speed of $10m/s$. It is subjected to a retarding force $F=0.1xJ/m$ during its travel from $x=20m$ to $x=30cm$. Its final kinetic energy will be

A

$450J$

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B

$475J$

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C

$250J$

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D

$275J$

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Solution

## The correct option is B $475J$Step1: Kinetic energy Kinetic energy is the energy created by an object as a result of its motion. When an object is set to accelerate, it is imperative that specific forces be applied. Work is required to apply force, and once the work is completed, the energy is transmitted to the object, causing it to move at a constant velocity.Step2: Given dataBlock of mass, $m=10kg$Regarding force, $F=0.1xJ/m$ and it is acted on the block from $x=20cmtox=30cm$Step3: Formula used$K.E=\frac{1}{2}m{v}^{2}\left[whereK.E=kineticenergy,m=mass,v=velocity\right]$Since the force F is a variable force, The work small amount of work done in moving the block to a small distance (dx) is given by,$dW=\int Fdx$Step4: Calculating kinetic energy$K.E.=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}K.E.=\frac{1}{2}×10×{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}K.E.=500J$Step5: Calculating work doneFurther, it is given to us that the block is subjected to a retarding force. Since the force F is a variable force, The work small amount of work done in moving the block to a small distance (dx) is given by,$dW=\int Fdx\phantom{\rule{0ex}{0ex}}\int dW={\int }_{20}^{30}0.1xdx\phantom{\rule{0ex}{0ex}}W=0.1{\int }_{20}^{30}xdx\phantom{\rule{0ex}{0ex}}W=0.1{\left[\frac{{x}^{2}}{2}\right]}_{20}^{30}\phantom{\rule{0ex}{0ex}}W=0.05\left[{30}^{2}-{20}^{2}\right]\phantom{\rule{0ex}{0ex}}W=0.05\left[900-400\right]\phantom{\rule{0ex}{0ex}}W=25J$Step6: Calculating final kinetic energy$K.E\left(Final\right)=K.E-W\phantom{\rule{0ex}{0ex}}K.E\left(Final\right)=500-25\phantom{\rule{0ex}{0ex}}K.E\left(Final\right)=475J$Hence, option B is the correct answer.

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