a constant force acts on an object of mass 5kg for a duration of 2 seconds it increases the object's velocity from 3m/s to 7m/s find the magnitude of the applied force now if the force where applied for a duration of 5 seconds what would be the final velocity of the object
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Answer:
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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