Question

Find the value of the constant force that is applied for 1 second on a 10 kg object increase the velocity of the object from 5 metre per second to 10 metre per second .

Answer 1

Answer

• The force will be 50 N

Explanation

Given

• Initial Velocity = 5 m/s
• Final Velocity = 10 m/s
• Time = 1 sec
• Mass = 10 kg

To Find

• The Force applied

Solution

• First find the acceleration of the body with the help of the first law of motion. Then on simply substitution of the values on the F = ma to find the force!!

✭ Acceleration of the body

→ v = u+at [First equation of motion]

• v = Final Velocity = 10 m/s
• u = Initial Velocity = 5 m/s
• a = Acceleration = ?
• t = Time = 1 sec

→ 10 = 5 + a×1

→ 10 = 5 + a

→ 10-5 = a

→ Acceleration = 5 m/s²

✭ Force Applied

→ F = ma

• m = Mass = 10 kg
• a = Acceleration = 5 m/s²

→ F = 10 × 5

→ Force = 50 N

Know More

Other two equations of motion

• s = ut+½at²
• v²-u² = 2as

Answer 2

Answer:

$\huge { \underline{ \large{ \rm{ \pink{Given:}}}}}$

• Initial Velocity (u) = 5m/s
• Final Velocity (v) = 10m/s
• Time (t) = 1sec
• Mass (m) = 10kg

$\huge{ \underline{ \large{ \rm{ \green{Find:}}}}}$

• Force Applied

$\huge { \underline{ \rm{ \large{ \orange{Solution:}}}}}$

• The second law of motion by Newton says that the force is equal to the change in momentum per change in the time. For a constant mass, force equals the mass times acceleration,F=m×a.

So,

$\boxed{ \sf{ \red{Force \: Applied = m \times a}}}$

Let's find the acceleration:-

• There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a).
• First Equation of Motion v=u+at.

$\to \large{ \sf{10 = 5 + a \times 1 }}$

$\to \large{ \sf{10 = 5 + a}}$

$\large \to{ \sf{a = 10 - 5 = 5}}$

$\boxed{ \therefore{ \purple{ \sf{Acceleration = 5}}}}$

☞Force applied:-

F = m × a

F = 10 × 5

F = 50

${ \boxed{ \therefore{ \blue{ \sf{Force = 50N}}}}}$

⇒Therefore, Force applied on object is 50N