Question

Newton's law of motion and prove that F=ma By applying force of any body of mass 2kg in rest it's velocity become 12m/s in 3 seconds find the value of applied force​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Correct\ Question \colon }}}}$

Derive the relation F = ma from Newton's law of motion. An object of mass 2 Kg in rest is acted upon by a force for three seconds, changing its velocity to 12 m/s. Find the magnitude of applied force.

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Newton's Second Law

The rate of change of momentum of an object depends upon the direction and magnitude of applied force.

$\large{\sf{F \propto \frac{dp}{dt}}}$

$\large{\leadsto \ \sf{F \propto \frac{d(mv - mu)}{dt}}}$

$\large{\leadsto \ \sf{F = k.m \frac{d(v - u)}{dt}}}$

$\large{\leadsto \ \sf{F = k.ma}} \\ \\ \sf{[If, k = 1]}$

$\\ \huge{\leadsto \ \underline{\boxed{\sf{F = ma}}}}$

From the Question,

• Mass of the object,m = 2 Kg

• Final Velocity,v = 12 m/s

• Time taken,t = 3s

To find

Force acting on the object

Using the Relation,

$\huge{\sf{v = u + at}}$

Putting the values,we get the acceleration of the object :

$\sf{12 = 0 + 3a} \\ \\ \leadsto \: \sf{3a = 12} \\ \\ \huge{ \leadsto \: \sf{a = 4 \: {ms}^{ - 2} }}$

Now,

$\huge{ \sf{f = ma}} \\ \\ \longrightarrow \: \sf{f = (4)(2)} \\ \\ \huge{\longrightarrow \: \tt{f = 8\: N}}$

The force acting on the object is 8 N

Answer 2

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

1) Derive the relation F = ma from Newton's law of motion.

2) An object of mass 2 Kg in rest is acted upon by a force for three seconds, changing its velocity to 12 m/s. Find the magnitude of applied force.

$\rule{200}{2}$

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

By newton's Second Law we know that :

$\Large \displaystyle {\boxed{\rm{F \: \propto \: \frac{dp}{dt}}}}$

$\Large \displaystyle \leadsto {\rm{F \: = \: \frac{d(mv \: - \: mu)}{dt}}}$

As,

$\Large \displaystyle \implies {\boxed{\sf{\underbrace{mv \: - \: mu}_{p}}}}$

Where, p is momentum

___________________

$\Large \displaystyle \leadsto {\rm{F \: = \: m\frac{d(v \: - \: u)}{dt}}}$

$\Large \displaystyle \leadsto {\rm{F \: = \: kma}}$

And k = 1 in S.I Units

$\LARGE \displaystyle {\boxed{\boxed{\bf{F \: = \: ma}}}}$

$\rule{200}{3}$

We are Given :

Final Velocity (v) = 12 m/s

Initial velocity (u) = 0 m/s

Mass (m) = 2 kg

Use First Equation of Motion

$\LARGE {\boxed{\boxed{\red{\bf{v \: = \: u \: + \: at}}}}}$

Put Values

⇒ 12 = 0 + (a)(3)

⇒3a = 12

⇒a = 12/3

$\Large \displaystyle \longrightarrow {\boxed{\rm{Acceleration \: = \: 4 \: ms^2}}}$

If Acceleration is 4 m/s²

Now use the above relation

Put Values

⇒ F = ma

⇒F = (2)(4)

⇒F = 8

$\LARGE \displaystyle {\boxed{\boxed{\orange{\mathbb{Force \: = \: 8 \: N}}}}}$