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A constant force acts on an object of mass 5 kg for a duration of 2s it increases the object's velocity from 3 m/s To 7 m/s. Find the magnitude of the applied force. Now if the force was applied for a duration of 5s what would be the final velocity of the object?
Given,
mass = 5kg
time = 2s
Initial velocity =
final velocity =
Let the acceleration be 'a'.
Therefore, $a\ =\ \frac{v\ -\ u}{t} \ =\ \frac{7\ -\ 3}{2} \ =\ 2\ m/s^{2}$
We know that, force is the product of mass and acceleration. Therefore, force = 5 x 2 = 10N.
Let the final velocity after 5s be 'v'.
Therefore,
$v\ =\ u\ +\ at\ =\ 3\ +\ 2\times 5=\ 13\ m/s$
Hence the final velocity of the object = 13m/s.
Updated on: 10-Oct-2022
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