Question

question in the attachment!!!solve the problem​

Answer 1

The initial kinetic energy,

$\sf{\longrightarrow K_i=\dfrac{1}{2}\,m(v_i)^2}$

$\sf{\longrightarrow K_i=\dfrac{1}{2}\cdot1 (2)^2}$

$\sf{\longrightarrow K_i=2\ J}$

• By work - energy theorem, the work done on the block when it moves along the patch is equal to change in kinetic energy.

$\sf{\longrightarrow K_f-K_i=W}$

$\displaystyle\sf{\longrightarrow K_f-K_i=\int\limits_{0.10}^{2.01}F_r\ dx}$

$\displaystyle\sf{\longrightarrow K_f-K_i=-k\int\limits_{0.10}^{2.01}\dfrac{1}{x}\ dx}$

$\displaystyle\sf{\longrightarrow K_f-K_i=-k\Big[\log x\Big]_{0.10}^{2.01}}$

$\displaystyle\sf{\longrightarrow K_f-2=-0.5\cdot\log\left(\dfrac{2.01}{0.10}\right)}$

$\displaystyle\sf{\longrightarrow K_f-2=-0.5\log(20.1)}$

$\displaystyle\sf{\longrightarrow K_f= 2-0.5\times3}$

$\displaystyle\sf{\longrightarrow\underline{\underline{K_f=0.5\ J }}}$

Answer 2

The change of energy is equal to work done so,

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$\sf{\dfrac{{{mv}^{2}_f}}{2}-\dfrac{{{mv}^{2}_f}}{2}=W}$

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Thus,

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$\sf{vf=\sqrt{\dfrac{2W}{m}+{v}^{2}_i}}$

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The work done is given by;

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$\displaystyle\sf{W=\int_{x_1}^{x_2}Fdx=-k\int_{x_1}^{x_2}\dfrac{dx}{x}}$

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$\sf{-k\:In\:\dfrac{x_2}{x_1}=-0.5\:In\:\dfrac{2.01}{0.1}=-1.5J}$

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Therefore,

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$\sf{v_f=\sqrt{\dfrac{2×(-1.5)}{1}+{2}^{2}=1m/s}}$

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The final kinetic energy;

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$\sf{K_f=\dfrac{{{mv}^{2}_f}}{2}=\dfrac{1×{1}^{2}}{2}=0.5J}$