
Let us assume that a vehicle of mass $20kg$ is moving with a velocity mentioned as $4m{{s}^{-1}}$. Calculate the magnitude of the force that is to be applied on the vehicle so that the vehicle will be having a velocity of \[1m{{s}^{-1}}\] after travelling a distance of \[20m\].
\[\begin{align}
& A.4N \\
& B.6N \\
& C.7.5N \\
& D.9.5N \\
\end{align}\]
Answer
559.2k+ views
Hint: The difference between the square of the final velocity and square of the initial velocity will be equivalent to twice the product of acceleration and the displacement of the body. Find the acceleration using this. The force can be found by taking the product of the mass and acceleration of the object. This will help you in answering this question.
Complete step by step answer:
According to the newton’s third equation of motion, we can write that,
\[{{V}^{2}}-{{U}^{2}}=2aS\]
Where \[V\] be the final velocity, \[U\] be the initial velocity, \[a\] be the acceleration of the vehicle and \[S\]be the displacement of the vehicle.
It has been already mentioned in the question that,
The displacement of the vehicle has been given as,
\[S=20m\]
The final velocity of the vehicle has been given as,
\[V=1m{{s}^{-1}}\]
The initial velocity of the vehicle has been mentioned as,
\[u=4m{{s}^{-1}}\]
Using this, the acceleration of the vehicle can be found by rearranging the equation,
\[\dfrac{{{V}^{2}}-{{U}^{2}}}{2S}=a\]
Substituting the values in it will give,
\[\begin{align}
& \dfrac{{{1}^{2}}-{{4}^{2}}}{2\times 20}=a \\
& \Rightarrow a=\dfrac{-3}{8}m{{s}^{-2}} \\
\end{align}\]
As the mass of the vehicle has been mentioned as,
\[m=20kg\]
Therefore the force applied on the vehicle can be found by the equation,
\[F=ma\]
Substituting the values in it will give,
\[F=20\times \dfrac{-3}{8}=7.5N\]
Therefore the magnitude of the force has been obtained as \[7.5N\].
So, the correct answer is “Option C”.
Note: A force will be a push or the pull experienced on a body which will cause the interaction of the object with another object. Whenever there will be an interaction between two bodies, there will be a force upon each of the bodies. The unit of the force has been found to be in newton.
Complete step by step answer:
According to the newton’s third equation of motion, we can write that,
\[{{V}^{2}}-{{U}^{2}}=2aS\]
Where \[V\] be the final velocity, \[U\] be the initial velocity, \[a\] be the acceleration of the vehicle and \[S\]be the displacement of the vehicle.
It has been already mentioned in the question that,
The displacement of the vehicle has been given as,
\[S=20m\]
The final velocity of the vehicle has been given as,
\[V=1m{{s}^{-1}}\]
The initial velocity of the vehicle has been mentioned as,
\[u=4m{{s}^{-1}}\]
Using this, the acceleration of the vehicle can be found by rearranging the equation,
\[\dfrac{{{V}^{2}}-{{U}^{2}}}{2S}=a\]
Substituting the values in it will give,
\[\begin{align}
& \dfrac{{{1}^{2}}-{{4}^{2}}}{2\times 20}=a \\
& \Rightarrow a=\dfrac{-3}{8}m{{s}^{-2}} \\
\end{align}\]
As the mass of the vehicle has been mentioned as,
\[m=20kg\]
Therefore the force applied on the vehicle can be found by the equation,
\[F=ma\]
Substituting the values in it will give,
\[F=20\times \dfrac{-3}{8}=7.5N\]
Therefore the magnitude of the force has been obtained as \[7.5N\].
So, the correct answer is “Option C”.
Note: A force will be a push or the pull experienced on a body which will cause the interaction of the object with another object. Whenever there will be an interaction between two bodies, there will be a force upon each of the bodies. The unit of the force has been found to be in newton.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

