Question

velocity of an object of mass 5kg increase from 3m/s to 7m/s on applying a force of 2s.find applied force​

Answer 1

Answer:

u=3 m/s

v=7 m/s

t=2s

m=5 kg

According to Second Law of Motion F=ma

= m(v−u)/t

= 5(7−3)/2

=10 N

Acceleration a= F/m

= 10/5

=2m/s^2

If we substitute the values in the equation v=u+at velocity can be calculated when the time for force is extended to 5s

v=3+(2×5)=12 m/s

Explanation:

thanks....

Answer 2

Answer:

$\large \pink{Expalation}$⍗

u=3 m/s

v=7 m/s

t=2s

m=5 kg

☞︎︎︎According to Second Law of Motion,

F=ma

= t__m(v−u) [Upon]

= 2_5(7−3) [Upon]

=10 N

Acceleration a= m_F [Upon] = 510 =2m/s (2)

If we substitute the values in the equation v=u+at velocity can be calculated when the time for force is extended to 5s

v=3+(2×5)=12 m/s

$\large \pink{Answer↝}$ 12m/s.

$\large \green{Mabelrose♡︎}$