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# A constant force F acts on a particle of mass 1 kg moving with a velocity v, for one second. The distance moved in that time is :A. $0$B. $\dfrac{F}{2}$C. $2F$D. $\dfrac{v}{2}$E. $v + \dfrac{F}{2}$

Last updated date: 17th Jun 2024
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Hint: If a constant force F is acting on a body of mass m, then the acceleration will be constant and given by $a = \dfrac{F}{m}$ as Newton’s second law of motion states that $F = ma$ .
If a body is moving with a constant acceleration $a$ , then from the equation of motion, we can say that $s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the displacement of that particle in time $t$ and $u$ is its initial velocity.

Complete step by step solution:
As given in the question that the force applied on the body of mass 1 kg is constant and equal to $F$ .
So, as we know that if a constant force F is acting on a body of mass m, then the acceleration will be constant and given by $a = \dfrac{F}{m}$ as Newton’s second law of motion states that $F = ma$ .
So, $a = \dfrac{F}{1} = F$
Now, it is given in the question that the body moves for 1 second i.e. $t = 1$
We also know that if a body is moving with a constant acceleration $a$ , then from the equation of motion, we can say that $s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the displacement of that particle in time $t$ and $u$ is its initial velocity.
Let the distance moved in that time be $s$
Then, from the equation of motion $s = ut + \dfrac{1}{2}a{t^2}$ ,
$s = v \times 1 + \dfrac{1}{2} \times F \times {1^2} = v + \dfrac{F}{2}$ (as $a = F$ and $u = v$)

$\therefore$The distance moved in that time is $v + \dfrac{F}{2}$. Hence, option (E) is the correct answer.

Note:
Remember that the equation of motion $s = ut + \dfrac{1}{2}a{t^2}$ is only applicable when the acceleration through which the body is moving is constant throughout the motion.
If the force applied on a body is constant throughout the motion then its acceleration will also be constant.