A body of mass $5$ kg rests on a rough horizontal surface of friction coefficient $0.2$ . The is pulled through a distance $10$ m by a horizontal force of $25 N$ . The kinetic energy acquired by it is

200 J

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150 J

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100 J

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50 J

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Solution

The correct option is B $150 J$
Friction force on body is $f = μ m g = 0.2 \times 5 \times 10 = 10 N$
no net force on body is $25 - 10 = 15 N$
So acceleration $a = 15 / 5 = 3 m s^{- 2}$
now using $v^{2} - u^{2} = 2 a S$ , also $u = 0$
so $v^{2} = 2 \times 3 \times 10 = 60$

KE= $\frac{m v^{2}}{2} = \frac{5 \times 60}{2} = 150 J$

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A body of mass 5 kg rests on a rough horizontal surface of friction coefficient 0.2. The is pulled through a distance 10 m by a horizontal force of 25N. The kinetic energy acquired by it is

The correct option is B 150JFriction force on body is f=μmg=0.2×5×10=10 Nno net force on body is 25−10=15 NSo acceleration a=15/5=3 ms−2now using v2−u2=2aS , also u=0 so v2=2×3×10=60KE=mv22=5×602=150 J

A body of mass $5$ kg rests on a rough horizontal surface of friction coefficient $0.2$ . The is pulled through a distance $10$ m by a horizontal force of $25 N$ . The kinetic energy acquired by it is