Solve the given problem:
A constant force acts on an object of mass \(4\) \(kg\) for a duration of \(6\) \(s\). It increases the object’s velocity from \(9\) \(m\ s^{-1}\) to \(11\) \(m\ s^{-1}\). Find the magnitude of the applied force. Now, if the force was applied for a duration of \(13\) \(s\).
What would be the final velocity of the object?
From the Newton's second law, We know that,
Final velocity \(=\) \(m\ s^{-1}\)
Important!
Note: Enter the values up to two decimal if needed