Question

Example 9.1 A constant force acts on an
object of mass 5 kg for a duration of
2 s. It increases the object's velocity
from 3 ms to 7 ms. Find the
magnitude of the applied force. Now, if
the force was applied for a duration of
55. what would be the final velocity of
the object?​

Answer 1

Solution

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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 13m/s