Question

A body of mass $5$ kg is moving with a velocity $20\,m{s^{ - 1}}$. If a force of $100$ N is applied on it for $10$ sec in the same direction as its velocity. What will be the velocity of the body $?$

Answer 1

Here, we can find the velocity of the body after the given time intervals by making use of Newton’s 2nd law which states that If an unbalanced external force acts on a body , the body accelerates and direction of acceleration is the same as direction of net force. Also, we have to use the 1st Kinematical equation of motion for uniformly accelerated motion.

Formulae used:
(1) Newton’s 2nd law, $F = ma$
Where, $F$ - External unbalanced force acting on the body, $m$ - mass of the body and $a$ - acceleration produced in the body.
(2) 1st Kinematical equation,
$v = u + at$
Where, $v$ - Final velocity of the body, $u$- Initial velocity of the body and $t$ - time interval for which the body is accelerated.

Complete step by step answer:
Here, we have to calculate the value of the final velocity of the body when it is accelerated by the force. Let us write the given information from the question,
$m = 5$ kg , $F = 100$ N,
$\Rightarrow u = 20\,m{s^{ - 1}}$ and $t = 10$ sec
First we calculate the acceleration of the body using Newton’s 2nd law and then put it into the 1st kinematical equation to get our desired answer. Using Newton’s 2nd law and putting the given values, we get
$F = ma$
$\Rightarrow a = \dfrac{F}{m}$
$\Rightarrow a = \dfrac{{100}}{5}$
$\Rightarrow a = 20\,m{s^{ - 2}} - - - - - - - (1)$
Now, Using 1st Kinematical equation and substituting the value of $a$ from above equation and given values, we get
$v = u + at$
$\Rightarrow v = 20 + 20 \times 10$
$\Rightarrow v = 20 + 200$
$\therefore v = 220\,m{s^{ - 1}}$

Thus, the velocity of the body after $10$ sec is $220\,m{s^{ - 1}}$.

Note: The kinematical equations of motion are valid only for one dimensional uniform accelerated motion. We should notice that the velocity of the body is directly proportional to the time when acceleration of the body is constant i.e. $v$ $\alpha$ $t$.