Question

A body of mass $40{\text{ }}kg$ falls down a vertical distance of $10\,m$ from rest. Calculate its final velocity:
(i) if it was subjected to a continuous downward thrust of $40{\text{ }}N$,
(ii) if it faces a resistance of $40{\text{ }}N$
(iii) if it is acted upon by a thrust of $40{\text{ }}N$ for a distance of $4\,m$ and then by a resistance of $40{\text{ }}N$ for the remaining distance.

Answer 1

Hint:We have to employ Newton’s Second Law of motion to find out the acceleration by substituting all the available values. Then, to find the final velocity we have to use Motion’s equations.

Complete step by step answer:
In the given question, the mass of the body$ = $$40{\text{ }}kg$.
Total distance to be covered ($s$)$ = $$10\,m$.
We have to find the final velocity $v$.
(i)Where a continuous downward thrust of $40{\text{ }}N$ is implied.
From Newton’s Second Law,
$ma = mg + 40$
Here $a$ is the acceleration and $g$ is the acceleration due to gravity.
$a = g + \dfrac{{40}}{m}$
Substituting the value of $g$$ = 10{\text{ }}\dfrac{m}{{{s^2}}}$ and $m = 40{\text{ }}kg$ we get,
$a = 10 + \dfrac{{40}}{{40}} \\
\Rightarrow a = 11 \\ $
Thus, the acceleration is $11{\text{ }}\dfrac{m}{{{s^2}}}$.
Now, from motion’s equation,
${v^2} - {u^2} = 2as$
Substituting the values of $a$,$s$ and $u = 0$ we get,
${v^2} - 0 = 2 \times 11 \times 10 \\
\therefore {v^2} = 220 $
With the help of the square root of the given value we find $v = 14.83$. So, the velocity of the body subjected to a continuous downward thrust of $40{\text{ }}N$is $14.83{\text{ }}\dfrac{m}{s}$.

(ii) When a resistance of $40{\text{ }}N$ is implied,
From Newton’s Second Law,
$ma = mg - 40$
Here $a$ is the acceleration and $g$ is the acceleration due to gravity.
$a = g - \dfrac{{40}}{m}$
Substituting the value of $g$$ = 10{\text{ }}\dfrac{m}{{{s^2}}}$ and $m = 40{\text{ }}kg$ we get,
$a = 10 - \dfrac{{40}}{{40}} \\
\Rightarrow a = 9 \\ $
Thus, the acceleration is ${\text{9 }}\dfrac{m}{{{s^2}}}$.
Now, from motion’s equation,
${v^2} - {u^2} = 2as$
Substituting the values of $a$,$s$ and $u = 0$ we get,
${v^2} - 0 = 2 \times 9 \times 10 \\
\therefore {v^2} = 180 $
After square root the value of $v$ is $13.42{\text{ }}\dfrac{m}{s}$.

(iii) A thrust of $40{\text{ }}N$ for a distance of $4\,m$ and then by a resistance of $40{\text{ }}N$ for the remaining distance.
Similarly,
$a = g + \dfrac{{40}}{m}$
Substituting the value of $g$$ = 10{\text{ }}\dfrac{m}{{{s^2}}}$ and $m = 40{\text{ }}kg$ we get,
$a = 10 + \dfrac{{40}}{{40}} \\
\Rightarrow a = 11 \\ $
Thus, the acceleration is $11{\text{ }}\dfrac{m}{{{s^2}}}$.
The final velocity of the body for $4\,m$distance be $v'$.
${v'^2} - {u^2} = 2as$
$ \Rightarrow {(v')^2} = 2 \times 11 \times 4$
So, the velocity $v' = 9.38{\text{ }}\dfrac{m}{s}$.
Now, for the rest $6{\text{ }}m$, $v'$becomes the initial velocity,
As it is acted upon a resistance of $40\;N$ it’s acceleration is ${\text{9 }}\dfrac{m}{{{s^2}}}$ from case-(ii)
${v^2} - {(v')^{}} = 2 \times 9 \times 6 \\
\therefore {v^2} = 88 + 108 = 196 $
Squaring the root we get the final velocity $v$ as $14{\text{ }}\dfrac{m}{s}$.

Note:For resistance the force is applied opposite to the direction of the body while the thrust is applied towards the direction of fall. It must be noted that in the last case-(iii) the velocity of the body gained in the first $4{\text{ }}m$ becomes the initial velocity of the last ${\text{6 }}m$. The acceleration due to gravity should be taken as $10{\text{ }}\dfrac{m}{{{s^2}}}$ for the ease of calculation.