Question

The mass of a certain object is 6kg. What is the force of that could have acted on it if its velocity increase from 3ms -1 to 13ms -1 during 5 seconds

Answer 1

Answer:-

Given:

Mass of an object = 6 kg.

Initial Velocity of the object (u) = 3 m/s

Final Velocity of the object (v) = 13 m/s.

Time taken = 5 s.

Acceleration produced = (v - u)/t.

→ a = (13 - 3)/5

→ a = (10)/5

→ a = 2 m/s².

We know that,

Force applied = mass of the object* Acceleration of the object.

→ F = 6 kg * 2 m/s²

→ F = 12 kg m/s²

→ F = 12 N.

Therefore, the force applied is 12 N.

Answer 2

Given:-

• Mass (m) of the particle = 6 kg
• Initial velocity (u) = 3 m/s
• Final velocity (v) = 13 m/s
• Time taken (t) = 5 seconds

To Find: Magnitude of the force exerted in the particle.

We know,

• F = ma &
• a = (v - u)/t

∴ F = m[(v - u)/t]

where,

• F = Force,
• a = Acceleration,
• v, u = Final and initial velocities respectively,
• t = Time taken.

∴ F = (6 kg)[(13 m/s - 3 m/s)/5s]

→ F = (6 kg)(2 m/s²)

→ F = 12 kg m/s²

→ F = 12 N

∴ Force exerted on the particle was 12 Newtons.