Question

Using second law of motion derive the relation between force and acceleration

Answer 1

f=ma

Step-by-step explanation:

According to second law of motion Force is directly proportional to rate of change of momentum.

so F is directly proportional to
`(dp)/(dt)`

we know that momentum P= MV

so

`f=d(mv)/dt=m(dv)/(dt)`

thus we get F=ma.

Answer 2

Let F be external force applied on the body in the direction of motion of the body for time interval
`\sf \Delta t`, the the velocity of a body of mass m changes from
`\sf v` to
`\sf v + \Delta v` i.e. change in momentum,
`\sf \Delta p = m \Delta v`.

According to Newton's second law :

`:\implies \sf F \propto (\Delta p)/(\Delta t) \\ \\ \\`

`:\implies \sf F = k \: (\Delta p)/(\Delta t) \\ \\ \\`

Where k is a constant of proportionality.

If limit
`\sf \Delta t \rightarrow 0,` then the term
`\sf (\Delta p)/(\Delta t)` becomes the derivative
`\sf (dp)/(dt)`.

Thus,

`:\implies \sf F = k \: (dp)/(dt) \\ \\ \\`

For a body of fixed mass (m), we have :

`:\implies \sf F = k (d(mv))/(dt) \\ \\ \\`

`:\implies \sf F = km \: (dv)/(dt) \\ \\ \\`

`:\implies \sf F = kma \\ \\ \\`

If v is fixed and m is variable then :

`:\implies \sf F = (kd(mv))/(dt) \\ \\ \\`

`:\implies \sf F = (kvdm)/(dt) \\ \\ \\`

because, k = 1 then :

`:\implies \sf F =(vdm)/(dt)`

Now, a unit force may be defined as the force which produces unit acceleration in a body of unit mass :]

So,

F = 1

m = 1

a = 1

k = 1

So,

`:\implies \underline{ \boxed{ \sf F = ma}}`