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# The net force on a rocket with a weight of $1.5 \times {10^4}$ N is $2.4 \times {10^4}$ N. About how much time is needed to increase the rocket’s speed from 12 $m/s$ to 36 $m/s$ near the surface of the Earth at take off? (Take $g = 10m/{s^2}$ ) (A) 15 s(B) 0.78 s(C) 1.5 s(D) 3.8 s

Last updated date: 13th Jun 2024
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Hint Firstly, find mass of the rocket from given values of weight and acceleration due to gravity. Then, use it to find acceleration of the rocket from the given value of net force and calculated mass of the rocket. Then put acceleration of the rocket into the equation of motion: $v = u + at$, along with other given values of initial and final speeds, to get how much time will be required.

Complete step by step solution
We are given that the weight of the rocket is $W = 1.5 \times {10^4}$ N and acceleration due to gravity is $g = 10m/{s^2}$ .
We know that the weight of the rocket (W) is equal to acceleration due to gravity (g) times the mass of the rocket (m). $\therefore W = mg$
$\Rightarrow m = \dfrac{W}{g}$
Putting in the values of W and g in the above equation,
$\Rightarrow m = \dfrac{{1.5 \times {{10}^4}}}{{10}} \\ \Rightarrow m = 1.5 \times {10^3}kg \\$
Hence, the mass of the rocket is $m = 1.5 \times {10^3}kg$.
Net force on the rocket (F) will be equal to acceleration of the rocket(a) times the mass of the rocket (m). We are given that the net force on the rocket is $F = 2.4 \times {10^4}$N.
$\Rightarrow F = ma \\ \Rightarrow a = \dfrac{F}{m} \\$
Substituting the values of net force on the rocket (F), which is given and mass of the rocket (m) that we calculated above in this expression for acceleration of the rocket gives us,
$\Rightarrow a = \dfrac{{2.4 \times {{10}^4}N}}{{1.5 \times {{10}^3}kg}} \\ \Rightarrow a = 1.6 \times 10\dfrac{{kgm{s^{ - 2}}}}{{kg}} \\ \Rightarrow a = 16m{s^{ - 2}} \\$
Now, we are given our initial speed of the rocket is 12 $m/s$ and the final speed of the rocket must become 36 m/s or we can say $u = 12m/s$ and $v = 36m/s$ .
Using the equation of motion : $v = u + at$ and substituting all the known values,
$\Rightarrow 36 = 12 + \left( {16 \times t} \right) \\ \Rightarrow 36 - 12 = 16t \\ \Rightarrow 24 = 16t \\ \Rightarrow t = \dfrac{{24}}{{16}} \\ \Rightarrow t = 1.5\sec \\$
Therefore, the time required to increase the rocket’s speed from 12 $m/s$ to 36 $m/s$ as per the given parameters will be 1.5 seconds.

Hence, Option (C) is correct.

Note: An alternate short-cut method:
Take, Impulse = Force × Time
$\Rightarrow mv - mu = Ft \\ \Rightarrow m(v - u) = Ft \\ \Rightarrow \dfrac{W}{g}(v - u) = Ft \\ \Rightarrow \dfrac{{W(v - u)}}{{gF}} = t \\$
Putting in all the known values in the above equation yields,
$\Rightarrow t = \dfrac{{1.5 \times {{10}^4} \times \left( {36 - 12} \right)}}{{10 \times 2.4 \times {{10}^4}}} \\ \Rightarrow t = \dfrac{{1.5 \times 24}}{{24}} \\ \Rightarrow t = 1.5\sec \\$