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  4. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of friction =0.1 . The change in kinetic energy in 10 s is: (g=10 m / s 2)

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Q. A body of mass $2\, kg$ initially at rest moves under the action of an applied horizontal force of $7\, N$ on a table with coefficient of friction $=0.1 .$ The change in kinetic energy in $10\, s$ is: $\left(g=10\, m / s ^{2}\right)$

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Solution:

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$a=\frac{\sum F}{m}=\frac{F_{ex}-\mu m g}{m}$
$=\frac{7 N -(0.1)(2 kg )\left(10 m / s ^{2}\right)}{2 kg }$
$=\frac{5 N }{2}=2.5 ms ^{-2}$
After $t=10 s$
$v=u+a t=0+(2.5)(10)=25 ms ^{-1}$
$\Delta K=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=\frac{1}{2}$
(2) $(25)^{2} J =625 J$.