The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 k g$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

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Solution

Initial $K E$ of the body,

$= \frac{1}{2} m v^{2} = \frac{1}{2} \times 25 \times 4 = 50 J$

Work done against resistive force,
$=$ Area between $F - x$ graph

$= \frac{1}{2} \times 4 \times 20 = 40 J$

Final $K E =$ Initial $K E -$ Work done against resistive force
$= 50 - 40 = 10 J$

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The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 4 m, its kinetic energy would be

Initial KE of the body, =12mv2=12×25×4=50J Work done against resistive force, = Area between F−x graph =12×4×20=40 J Final KE= Initial KE− Work done against resistive force =50−40=10 J

The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 k g$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

Initial $K E$ of the body,

$= \frac{1}{2} m v^{2} = \frac{1}{2} \times 25 \times 4 = 50 J$

Work done against resistive force,
$=$ Area between $F - x$ graph

$= \frac{1}{2} \times 4 \times 20 = 40 J$

Final $K E =$ Initial $K E -$ Work done against resistive force
$= 50 - 40 = 10 J$