A body of mass $10$ kg at rest is acted upon simultaneously by two forces $4$ N and $3$ N at right angles to each other. The kinetic energy of the body at the end of $10$ sec is?

100

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300

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50

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125

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Solution

The correct option is D $125$ J
Since it is acted upon simultaneously by two forces $4 N$ and $3 N$ at right angles, net force $= \sqrt{4^{2} + 3^{2}} = 5 N$
The mass is given as $10 K g \to a = \frac{F}{m} = \frac{5}{10} = 0.5 \frac{m}{s^{2}}$
we calculate the kinetic energy $K . E = \frac{1}{2} m v^{2} = \frac{1}{2} m (a t)^{2}$
$K E = \frac{1}{2} 10 \times 0.52 \times 10 s = 125 J$

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A body of mass 10kg at rest is acted upon simultaneously by two forces 4N and 3N at right angles to each other. The kinetic energy of the body at the end of 10 sec is?

The correct option is D 125 J Since it is acted upon simultaneously by two forces 4N and 3N at right angles, net force =√42+32=5NThe mass is given as 10Kg→a=Fm=510=0.5ms2we calculate the kinetic energy K.E=12mv2=12m(at)2KE=1210×0.52×10s=125J

A body of mass $10$ kg at rest is acted upon simultaneously by two forces $4$ N and $3$ N at right angles to each other. The kinetic energy of the body at the end of $10$ sec is?