Define the force state newton second law of motion ans derive
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According to the Newton's 2ndLaw of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force and in the direction of force. It means that the linear momentum will change faster when a bigger force is applied.
Derivation
Consider a body of mass ‘m’ moving with velocity v.
The linear momentum of a body is given by:
p = mv
Now According to Newton’s 2nd Law of Motion:
Force is directly proportional to rate of change of momnetum, that is
F α dp/dt
F = k dp/dt
F = k d(mv)/dt
F = k md(v)/dt
F = k ma
Experimentally k =1
F = k ma
Which is the required equation of force.
Derivation
Consider a body of mass ‘m’ moving with velocity v.
The linear momentum of a body is given by:
p = mv
Now According to Newton’s 2nd Law of Motion:
Force is directly proportional to rate of change of momnetum, that is
F α dp/dt
F = k dp/dt
F = k d(mv)/dt
F = k md(v)/dt
F = k ma
Experimentally k =1
F = k ma
Which is the required equation of force.
Answered by
1
Hey Chika ❤
Here's ur answer-
![\underline{second \: law \: of \: motion} \underline{second \: law \: of \: motion}](https://tex.z-dn.net/?f=+%5Cunderline%7Bsecond+%5C%3A+law+%5C%3A+of+%5C%3A+motion%7D)
It states that the rate of change of momentum of a body is directly proportional to the applied force, and the best place in the direction in which the force acts.
![force \: = \: mass \times acceleration \\ or \\ f \: = \: m \times a force \: = \: mass \times acceleration \\ or \\ f \: = \: m \times a](https://tex.z-dn.net/?f=force+%5C%3A++%3D++%5C%3A+mass+%5Ctimes+acceleration+%5C%5C+or+%5C%5C+f+%5C%3A++%3D++%5C%3A+m+%5Ctimes+a)
________________________
✴ Consider a body of mass 'm' having an initial velocity 'u'.
✴The initial momentum of the body will be 'mu'
-----------------
✴ Suppose a force 'F' acts on this body for time 't' and causes final velocity to become 'v'.
✴The final momentum of this body will be 'mv'.
Now, Change in momentum of this body is (mv-mu) and the time taken for this change is 't' .
So, according to Newton's second law of motion:-
(in the attachment 1)
But [(v-u)/t ] represents change in velocity with time which is known as acceleration 'a'.
So we write 'a' instead [(v-u)/t ] in the attachment 1. and we get,
(check in the attachment 2)
Thus, the force acting on a body is directly proportional to the product of mass and acceleration.
Which gives:-
![f = m \times a f = m \times a](https://tex.z-dn.net/?f=f+%3D+m+%5Ctimes+a)
The Newton's second law of motion gives us a relationship between Force and acceleration.
Hope it helps ✔
Here's ur answer-
It states that the rate of change of momentum of a body is directly proportional to the applied force, and the best place in the direction in which the force acts.
________________________
✴ Consider a body of mass 'm' having an initial velocity 'u'.
✴The initial momentum of the body will be 'mu'
-----------------
✴ Suppose a force 'F' acts on this body for time 't' and causes final velocity to become 'v'.
✴The final momentum of this body will be 'mv'.
Now, Change in momentum of this body is (mv-mu) and the time taken for this change is 't' .
So, according to Newton's second law of motion:-
(in the attachment 1)
But [(v-u)/t ] represents change in velocity with time which is known as acceleration 'a'.
So we write 'a' instead [(v-u)/t ] in the attachment 1. and we get,
(check in the attachment 2)
Thus, the force acting on a body is directly proportional to the product of mass and acceleration.
Which gives:-
The Newton's second law of motion gives us a relationship between Force and acceleration.
Hope it helps ✔
Attachments:
![](https://hi-static.z-dn.net/files/d81/85389b593c0e317673e368ca7e232ecb.jpg)
![](https://hi-static.z-dn.net/files/d86/2f28333996c586fefbd6188a9f0f45e8.jpg)
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