When a constant force acts on an object with a mass of 8 kg for a duration of 3 s. it increases the object's velocity from 4 m/s to 6 m/s . What is the magnitude of the applied force?

A) 5.33 N B) 4.33 N C) 6.33 N D) 3.33 N

Answer & Solution

Correct Answer: Option A

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Question posted by Examveda

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2 Comments

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Udochukwu Peace 3 years ago

Force = mass × acceleration.
Acceleration= (final velocity - initial velocity)/ time
Final velocity= 6m/s; initial velocity= 4m/s
6-4=2m/s;
Acceleration= 2/3= 0.3333
Force= 8×0.3333= 5.33N

Namgay Jr 6 years ago

Force = mass × acceleration
For v1, a1 = 4/3 m/s^2
For v2, a2 = 6/3 m/s^2
So, magnitude of applied force F = m×a1 - m×a2
= 5.33 N

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When a constant force acts on an object with a mass...

When a constant force acts on an object with a mass of 8 kg for a duration of 3 s. it increases the object's velocity from 4 m/s to 6 m/s . What is the magnitude of the applied force?

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