Question

A constant force acts on an object of mass 2 Kg for 10 s and increases its velocity from 5 m/s to 10 m/s. the magnitude of applied force will be

Answer 1

Answer :

Mass of object = 2kg

Initial velocity = 5m/s

Final velocity = 10m/s

Time interval = 10s

We have to find magnitude of applied force$.$

As per Newton's second law of motion, Force is defined as the rate of change of linear momentum.

• F = m (v - u) / t

➔ F = m (v - u) / t

➔ F = 2 (10 - 5) / 10

➔ F = 2 (5) / 10

➔ F = 10/10

➔ F = 1N

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Answer 2

✏ Given :-

➠ Mass = 2kg.

➠ Initial Velocity = 5m/s

➠ Final Velocity = 10m/s

➠ Time Taken = 10 seconds.

✏ To Find :-

➠ Magnitude Of Force.

✏ Solution :-

We know that,

F = ma

Where,

F = Force,

M = Mass,

A = Acceleration.

We also know that,

A = Initial Velocity - Final Velocity/Time.

Put the value of a.

So, The Formula :-

F = m(v-u)/t

F = 2(10-5)/10

F = 10/10

F = 1N.

Hence, The Force = 1N.

⭐ BASIC DEFINITIONS :-

• Force is the pull or push upon an object.

• Velocity is the quantity which specifies both the direction of motion and speed.

• Acceleration of a body is defined as the rate of change of velocity.