Question

A constant force acts on an object of mass 5 Kg for a duration of $2$ seconds. It increases the objects $v = u + at$velocity from $3\,m{s^{ - 1}}$ to $7\,m{s^{ - 1}}$ . Find the magnitude of the applied force. Now if the tax were applied for a duration of $5$seconds, what would be the final velocity of the object $?$

Answer 1

Hint
In the question, velocity of the objects is given, but we have to assume that initial velocity and final velocity. From the mass and acceleration of the object, we have to find the force of the object.
The expression of finding acceleration is $a = \left( {\dfrac{{v - u}}{t}} \right)$
Where, $a$ be the acceleration, $u$ be the initial velocity, $v$ be the final velocity and $t$ be the time $($Here time is measured in seconds$)$.
The expression of finding final velocity is $v = u + at$.
According to the newton second law of motion the expression of finding force is $f = ma$
Here, $m$ be the mass and $f$ be the force.

Complete step by step solution
From the question, we know that-
Mass of the object $m = 5\,kg$
Initial velocity $u = 3\,m{s^{ - 1}}$ and final velocity $v = 7\,m{s^{ - 1}}$ .
Time ${t_1} = 2 s$ and ${t_2} = 5s$.
Acceleration $a = \left( {\dfrac{{v - u}}{t}} \right)\,\,m{s^ - }^2...........\left( 1 \right)$
Apply the value of $u$and $v$in the equation $\left( 1 \right)$,
$a = \left( {\dfrac{{7 - 3}}{2}} \right)\,m{s^{ - 2}}$.
Hence, acceleration $a = 2\,m{s^{ - 2}}$.
Magnitude of the Applied Force:
Force = mass x acceleration $(f = ma)$
$f = 5\,kg \times 2\,m{s^{ - 2}}$
$f = 10\,N$
From the Question, we know that force is applied for $5$seconds, So we find the final velocity after $5$seconds
$v = u + at$
$v = 3\;m{s^{ - 1}} + (2\;m{s^{ - 2}} \times 5\;s)$
Perform the arithmetic operations, we get-
$v = 13\,m{s^{ - 1}}$
Hence, the final velocity of the object after 5 seconds is $13\,m{s^{ - 1}}$.

Note
In this question, the final velocity of the object is increased, so we assume the value of final velocity is high compared to initial velocity and the mass is applied continuously for $5$ seconds. So we have to find the force acting on $5$ seconds.