Question

A force acts on a particle of mass 200g the velocity of the particle changes from 15ms-2 to 25ms-2 in2.5s assuming the force to be constant ,the magnitude of force is

Answer 1

Answer:

0.8 N

Explanation:

Given:

• Mass of particle (m) = 200 g = 200/1000 = 0.2 kg
• Initial velocity (u) = 15 m/s
• Final velocity (v) = 25 m/s
• Time (t) = 2.5 s

To find:

• Magnitude of force (F) = ?

Solution:

Let's analyze this situation !

There's a particle of mass 0.2 kg whose velocity changes from 15 m/s to 25 m/s in 2.5 seconds. Assuming that the force acting on it is constant, we have to find the magnitude of force.

Let's find now !

By Newton's 2nd law, we comes to know that Force (F) = mass (m) × acceleration (a).

Here, we are not having the value of acceleration, so let's find it first.

$\:\:\:\:\:\:\:$ Acceleration = (v - u)/t

$\implies\:\sf{a\:=\:\dfrac{25\:-\:15}{2.5}}$

$\implies\:\sf{a\:=\:\dfrac{10}{2.5}}$

$\implies\:\sf{a\:=\:4\:m/s^2}$

Now, we know the acceleration and mass let's find the force !

$\implies$ F = 0.2 × 4

$\implies$ F = 0.8 N

${\underline{\underline{\sf{\large{\therefore\:Force\:is\:0.8\:Newton}}}}}$

Answer 2

GiveN :-

• Mass of particle = 200g = 0.2 Kg

• Initial velocity of particle = 15 m/s

• Final velocity of particle = 25 m/s

• Time = 2.5 s

To FinD :-

• Magnitude of the force acted on particle

SolutioN :-

Firstly we have to find acceleration of the body

$:\implies \boxed{ \bf \red{Acceleration = \frac{v - u}{t}}} \\ \\:\implies \sf a = \frac{25 - 15}{2.5} \\ \\:\implies \sf a = \frac{10}{2.5} \\ \\:\implies \boxed{\sf a = 4 \: {ms}^{ - 2}}$

Now Force on the particle is

$:\implies \boxed{ \bf \green {Force = Mass \times Acceleration}} \\ \\:\implies \sf f = 0.2 \times 4 \\ \\:\implies\boxed{ \sf f = 0.8 \: N}$

$\large\therefore \underline{\bf \blue{ Magnitude\: of\: Force \:is\:0.8\: N} }$