Question

an object initially moving with velocity 13m/s is subjected to a force 'f' which increases it's velocity upto 33m/s in 5sec. if mass of the object is 0.005kg. find force in standard form.​

Answer 1

GIVEN :-

• Initial velocity (u) = 13m/s
• Final velocity (v) = 33m/s
• Time taken (t) = 5 s
• Mass of object (m) = 0.005kg

TO FIND :-

• Force applied on the object.

SOLUTION :-

By 1st Kinematical equation ,

★ v = u + at

We have ,

• v = 33m/s
• u = 13m/s
• t = 5 s
• a = ?

Putting values in the equation,

→ 33 = 13 + a(5)

→ 33 - 13 = 5a

→ 20 = 5a

→ a = 20/5

→ a = 4m/s²

Hence , acceleration of the body is 4m/s².

_______________

By 2nd Law of Motion ,

★ F = ma

We have ,

• m = 0.005kg
• a = 4m/s²

Putting values ,

→ F = 0.005 × 4

→ F = 0.02 kg-m/s²

kg-m/s² is known as Newton (N) .

Hence , force applied of the object is 0.02N.

MORE TO KNOW :-

★ 2nd Kinematical equation :-

• s = ut + (1/2) at²

★ 3rd Kinematical equation :-

• v² = u² + 2as

★ Kinematical equations are only applicable when acceleration is constant.

Answer 2

$v = u + at \\ 33 = 13 + a(5) \\ 33 - 13 = 5a \\ a = \frac{20}{5} = {4ms}^{ - 2 } \\ acceleration = a = 4 {ms}^{ - 2}$

$f = ma \\ f = 0.005 \times 4 \\ f = 0.02 {kgms}^{ - 2} \\ hence \: force = 0.02n$