Question

a force of 2N acts for 5 second on a particle of mass 0.5Kg initially at rest calculate the distance moved by the particles for these 5 seconds​

Answer 1

AnswEr :

The distance moved by the particle in these five seconds is 50 m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Here,

$\textsf{ \textbf{GivEn}}\begin{cases}\sf \: \: Force = 2 \: N \\ \\ \sf \: \: Required \: time = 5 \: s \\ \\ \sf \: \: Mass \: of \: the \: Particle = 0.5 \: kg \\ \\ \sf \: \: Initial \: velocity = 0 \: ms {}^{ - 1} \end{cases}$

Firstly, we'll calculate the Acceleration & then by using the equations of motion we'll calculate the distance moved by the particle in these five seconds.

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \sf \: Acceleration = \dfrac{F}{m}$

$\dashrightarrow \sf \: Acceleration = \dfrac{2}{0.5}$

$\dashrightarrow \sf \: { \boxed{ \frak{\pink{Acceleration = 4 \: ms {}^{ - 2}}}}} \: \bigstar$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

By using Equations of Motion :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$: \implies \sf \: s = ut + \dfrac{1}{2}at^2$

$: \implies \sf \: s = 0 + \dfrac{1}{2} \times 4 \times 5^2$

$: \implies \sf \: s = 2 \times 5^2$

$: \implies \sf \:{ \boxed{ \purple{ \frak{ s = 50 \: m}}}} \: \bigstar$

$\therefore$The distance moved by the particle in these five seconds is 50 m.

⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀More to know :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Basically, there are 3 equations of motion. They're :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\;\;\;\;\sf\bullet\;v = u + at$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\;\;\;\;\sf\bullet\;v^2 = u^2 + 2as$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\;\;\;\;\sf\bullet\; s = ut + \dfrac{1}{2}at^2$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Answer 2

GiveN :

• Force (F) = 5 N
• Time (t) = 5s
• Mass (m) = 0.5 kg
• Initial velocity (u) = 0 m/s

To FinD :

• Distance Travelled

SolutioN :

$\implies$ F = ma

⠀⠀⠀⠀⠀⠀

$\implies$ 2 = 0.5 * a

⠀⠀⠀⠀⠀⠀

$\implies$ a = 2/0.5

⠀⠀⠀⠀⠀⠀

$\implies$ a = 20/5

⠀⠀⠀⠀⠀⠀

$\implies$ a = 4 $\sf{ms^{-2}}$

________________________

Use 2nd equation of Kinematics :

$\implies$ S = ut + ½ at²

⠀⠀⠀⠀⠀⠀

$\implies$ S = 0 + ½ (4) (5)²

⠀⠀⠀⠀⠀⠀

$\implies$ S = ½ * 4 * 25

⠀⠀⠀⠀⠀⠀

$\implies$ S = 2 * 25

⠀⠀⠀⠀⠀⠀

$\implies$ S = 50 $\sf{m}$