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

Question

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 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/23f99dd89cea62bc1aa30546c207f8e6-.png" /> (A) 50 J (B) 40 J (C) 20 J (D) 10 J

Tardigrade Admin · Accepted Answer

Initial K.E. of the body =(1/2) m v2 =(1/2) × 25 × 4=50 J Work done against resistive force = Area between F-x graph =(1/2) × 4 × 20=40 J Final K.E. = Initial K.E. - Work done against resistive force =50-40=10 J

Q. 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

50 J

40 J

20 J

10 J

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
$= 50 - 40 = 10 J$

Q. 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 25kg and initial velocity is 2m/s. When the distance covered by the body is 4m, its kinetic energy would be