Question

Figure shows the position-time graph of a particle of mass

4 kg. Let the force on the particle for $t < 0,0 < t < 4s,tF_{2}$ $> 4sbeF_{1},$ and$F_{3}$respectively. Then:

• A. $F_{1} = F_{2} = F_{3} = 0$
• B. $F_{1} > F_{2} = F_{3}$
• C. $F_{1} > F_{2} > F_{3}$
• D. $F_{1} < F_{2} < F_{3}$

Answer 1

Answer: (A)

$F_{1} = F_{2} = F_{3} = 0$

Answer 2

For $t < 0$and $t > 4s$the position of the particle is not chaining i.e. the particle is at rest so no force is acting on the particle at these intervals.

For $0 < t < 4s$the position of the particle is continuously chaining. As the position - time graph is s straight line, the motion of the particle is uniform, so acceleration, a=0 Hence no force acts on the particle during this interval also.