Physics, asked by pankajbhardwaj78, 7 months ago

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

Answers

Answered by ruhi08
7

Answer:

Given,

mass=5kg

t1 =2s

Initial velocity u=3m/s

Final velocity v=7m/s

t2 = 5s

So,

a = (v - u) / t = 7-3 / 2 = 2m/s²

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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Answered by mrsumitsprasad
0

Mass_ 5kg

Initial velocity (u)_3m/s

Final velocity (v)_7m/s

Time_ 2s

Acceleration(a)= v- u/t

=7-3/2

=2m/s^

Applied Force = Mass X Acceleration

= 5x2

= 10 N

Now, new time(t) =5s

Final velocity = u + at

= 3 + 2*5

= 3 + 10

= 13 m/s Ans

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