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# 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 kinetic friction = 0.1. Compute the work done by the applied force in 10 s,

Q2 (a)  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 kinetic friction = 0.1. Compute the  work done by the applied force in 10 s,

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Using Newton's law we can write :

$a\ =\ \frac{F}{m}$

$=\ \frac{7}{2}\ =\ 3.5\ m/s^2$

The frictional force is given by :

$f\ =\ \mu mg$

$=\ 0.1\times 2\times 9.8\ =\ 1.96\ N$

Its direction will be opposite of the direction of the motion. Thus acceleration produced will be negative.

$a\ =\ \frac{-1.96}{2}\ =\ -0.98\ m/s^2$

Thus the net acceleration is  =   3.5  -   0.98  =   2.52 m/s2.

The total distance travelled is given by :

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or                                                  $=\ 0\ +\ \frac{1}{2}(2.52)10^2\ =\ 126\ m$

Hence the work done by applied force is given by :

$W\ =\ F.s\ =\ 7\times 126\ =\ 882\ J$

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