Question

a constant force acts on an object of mass 5 kg for a duration of 2 seconds. It increases the speed from 3 metre per second to 7 metre per second. Find force now if the force were applied for 5 seconds what will be the final velocity​

Answer 1

Answer:

Given,

mass=5kg

t1=2s

Initial velocity u=3m/s

Final velocity v=7m/s

t2=5s

So,

Let the Force be F

Let the acceleration be a

So,

 a=t(v−u)=2(7−3)=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 

Explanation:

hope it help you!

Answer 2

Answer:

Explanation:

By Newton's 2nd law of motion, the net force on an object equals its mass times its acceleration.

What is the acceleration on this object?  Acceleration is defined to be change in velocity over change in time.  This object changes velocity from 3 m/s to 7 m/s in 2 seconds.  Hence:

a = (vf-vi)/Δt = (7 m/s - 3 m/s)/(2 s) = 4/2 = 2 m/s2

Then, Fnet = ma = (5 kg)(2 m/s2) = 10 N

if the force was applied for a duration of 5s,what would be the final  velocity of the body..

hence, we need to final velocity of the object

here,

mass=m=5kg

time=t=5sec

force=f=10N

initial velocity=u=3m/sec final velocity=v=?

f=m*(V-U)/t

10=5*(V-3)/5

10*5=5*(V-3)

50/5=V-3

10=V-3

10+3=V

V=13m/sec

hope this is heplful for you...

thank you...