Question

79. A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object? ​

Answer 1

Answer:

velocity is 13m/s

Explanation:

mass=5kg

t  

1

​

=2s

Initial velocity u=3m/s

Final velocity v=7m/s

t  

2

​

=5s

So,

Let the Force be F

Let the acceleration be a

So,

a=  

t

(v−u)

​

=  

2

(7−3)

​

=2m/s  

2

 

 

So the magnitude of the applied force is 10N

And the final velocity after 5s is v  

So,

v=u+at

v=3+2×5

v=13m/s

The final velocity after 5s is 13m/s

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