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Question: A body is moving with a speed of \( 1m/s \) and a constant force \( F \) is needed to stop it in a d...

A body is moving with a speed of 1m/s1m/s and a constant force FF is needed to stop it in a distance xx . If the speed of the body is 3m/s3m/s the force needed to stop it in the same distance xx will be :
(A) 1.5F\left( A \right){\text{ 1}}{\text{.5F}}
(B) 3F\left( B \right){\text{ 3F}}
(C) 6F\left( C \right){\text{ 6F}}
(D) 9F\left( D \right){\text{ 9F}}

Explanation

Solution

Here in this question we have to find the force required to stop and from this, we will use the equation v2=u2+2as{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
v2=u2+2as{v^2} = {u^2} + 2as
Here, vv , will be the final velocity,
uu , will be the initial velocity,
ss , 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
v2=u2+2a1s\Rightarrow {v^2} = {u^2} + 2{a_1}s
And on substituting the values. We will get the equation as
02=32+2a1x\Rightarrow {0^2} = {3^2} + 2{a_1}x
And on taking the constant term one side, we get
a1=322x\Rightarrow {a_1} = \dfrac{{{3^2}}}{{2x}}
And on solving the above equation, we get
a1=92x\Rightarrow {a_1} = \dfrac{9}{{2x}}
Similarly, for the first case, the equation will become
v2=u2+2a2s\Rightarrow {v^2} = {u^2} + 2{a_2}s
And on substituting the values. We will get the equation as
02=12+2a2x\Rightarrow {0^2} = {1^2} + 2{a_2}x
And on taking the constant term one side, we get
a2=122x\Rightarrow {a_2} = \dfrac{{{1^2}}}{{2x}}
And on solving the above equation, we get
a2=12x\Rightarrow {a_2} = \dfrac{1}{{2x}}
So on equating bot the acceleration, we will get the equation as
a12x9=a22x1\Rightarrow \dfrac{{{a_1} \cdot 2x}}{9} = \dfrac{{{a_2} \cdot 2x}}{1}
Since the like term will cancel each other, so we get
a1=9a2\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 (D)\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.