$A$ constant force acts on an object of $5 \ kg$ for a period of $2 \ s$. It increases the velocity of the object from $3 \ m s^{-1}$ to $7 \ m s^{-1}$. Find the magnitude of the applied force. Now,if the force were applied for a period of $5 \ s$,what would be the final velocity of the object?
(D) Given: Mass $m = 5 \ kg$,initial velocity $u = 3 \ m s^{-1}$,final velocity $v = 7 \ m s^{-1}$,time $t = 2 \ s$.
Using Newton's second law of motion,$F = m \times a = m \times \frac{(v - u)}{t}$.
Substituting the values: $F = \frac{5 \times (7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \ N$.
Now,for the second part,the force $F = 10 \ N$ is applied for $t' = 5 \ s$.
The acceleration $a = \frac{F}{m} = \frac{10}{5} = 2 \ m s^{-2}$.
Using the first equation of motion,$v' = u + a \times t'$.
$v' = 3 + (2 \times 5) = 3 + 10 = 13 \ m s^{-1}$.
Using Newton's second law of motion,$F = m \times a = m \times \frac{(v - u)}{t}$.
Substituting the values: $F = \frac{5 \times (7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \ N$.
Now,for the second part,the force $F = 10 \ N$ is applied for $t' = 5 \ s$.
The acceleration $a = \frac{F}{m} = \frac{10}{5} = 2 \ m s^{-2}$.
Using the first equation of motion,$v' = u + a \times t'$.
$v' = 3 + (2 \times 5) = 3 + 10 = 13 \ m s^{-1}$.
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