Question

A body is moving with a speed of $ 1m/s $ and a constant force $ F $ is needed to stop it in a distance $ x $ . If the speed of the body is $ 3m/s $ the force needed to stop it in the same distance $ x $ will be :
 $ \left( A \right){\text{ 1}}{\text{.5F}} $
 $ \left( B \right){\text{ 3F}} $
 $ \left( C \right){\text{ 6F}} $
 $ \left( D \right){\text{ 9F}} $

Answer 1

Hint: Here in this question we have to find the force required to stop and from this, we will use the equation $ {v^2} = {u^2} + 2as $ . From this, we will get the ratio of the acceleration and by equating the acceleration we will get to the answer.

Formula used
 $ {v^2} = {u^2} + 2as $
Here, $ v $ , will be the final velocity,
 $ u $ , will be the initial velocity,
 $ s $ , will be the displacement.

Complete step by step answer:
So in this question, we have the initial velocity given and in each of the cases the final velocity will be zero. So by using the formula for the first case the equation will become
 $ \Rightarrow {v^2} = {u^2} + 2{a_1}s $
And on substituting the values. We will get the equation as
 $ \Rightarrow {0^2} = {3^2} + 2{a_1}x $
And on taking the constant term one side, we get
 $ \Rightarrow {a_1} = \dfrac{{{3^2}}}{{2x}} $
And on solving the above equation, we get
 $ \Rightarrow {a_1} = \dfrac{9}{{2x}} $
Similarly, for the first case, the equation will become
 $ \Rightarrow {v^2} = {u^2} + 2{a_2}s $
And on substituting the values. We will get the equation as
 $ \Rightarrow {0^2} = {1^2} + 2{a_2}x $
And on taking the constant term one side, we get
 $ \Rightarrow {a_2} = \dfrac{{{1^2}}}{{2x}} $
And on solving the above equation, we get
 $ \Rightarrow {a_2} = \dfrac{1}{{2x}} $
So on equating bot the acceleration, we will get the equation as
 $ \Rightarrow \dfrac{{{a_1} \cdot 2x}}{9} = \dfrac{{{a_2} \cdot 2x}}{1} $
Since the like term will cancel each other, so we get
 $ \Rightarrow {a_1} = 9{a_2} $
Here, the mass will be constant.
Therefore, the required force will be nine times greater than the earlier force.
Hence, the option $ \left( D \right) $ is correct.

Note:
For solving such types of questions we have to play with the equations and for this, the concept should be clear. And also we have to first check the units whether it is correct or needed to be changed then only we should proceed further for solving it.