Question

Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on
applying a force continuously for 2 s. Find out the force applied. If the
duration for which force acts is extended to 5 s, what will be the velocityof the object then?
u = 3m?/s
V = 7 m/s
t = 2 s
m = 5 kg​

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 13

Answer 2

Explanation:

do this queation by the 1st equation of motion that is v=u+at so first we dont know the acceleration we hace to find it out so,

a-change in velocity÷ time taken

a- 7÷2

a=14

so now velocity=

v=u+at

v=3+28 (14*5)

v=70 m/s

therefore the velocity of the object will be 70 m/s