Question

constant force acts on an object of mass 5kg for a duration of 2 s.It increases the objects velocity from 3m/s to 7m/s.Find the magnitude of applied force. *

5N

10N

2N

15N

​

Answer 1

Answer:

Given,

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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Answer 2

Answer:

10 N

Explanation:

Mass of the object = 5 kg

Initial velocity (u) = 3 m/s

Final velocity (v) = 7 m/s

Time taken (t) = 2 s

Acceleration = v-u/t

= (7 m/s - 3 m/s)/2 s

= 4 m/s / 2 s

= 2 m/s²

Force = ma

= 5 kg × 2 m/s²

= 10 kg m/s²

= 10 N

Therefore, the magnitude of applied force is 10 N.