Question

A constant force acts on an object of mass 5.5 kg for a duration of 2.5 s. It increases the object’s velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force.

Answer 1

Answer:

Let the Force be F

mass = 5.5kg

time(t) = 2.5 s so a = (v - u)/t = (7-3)/2 = 2 m/s²

initial velocity(u) = 3 m/s final velocity(v) = 7 m/s

so let the acceleration be a

So the magnitude of the applied force is 10 N

and the final velocity after 5 s is v

so v = u + at v=3+2.5x5.5

v = 16.75m/s

The final velocity after 5.5 s is 16.75 m/s

Answer 2

Explanation:

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