Question

An object of mass 1 kg was moving at a speed of $10{ms}^{-1}$. Constant force acts for 4s on the object and the speed of object was $2{ms}^{-1}$​ in the opposite direction after the applied force was removed. The force acting on the object was

• A. The magnitude of force was 3N
• B. The magnitude of force was -3N
• C. The direction of force was same as the direction of initial velocity
• D. The direction of force was opposite to the direction of initial velocity

Answer 1

Answer: (A), (D)

The direction of force was opposite to the direction of initial velocity

Let initial velocity be $u$, final velocity be $v$, and time taken be $t$.
Initial velocity, $u=10{ms}^{-1}$​
Final velocity, $v=-2{ms}^{-1}$​ (since this velocity is in opposite direction of the initial velocity which is taken as positive)
Time taken, $t=4s$
Let $a$ be the change in acceleration.
$a=\frac{v-u}{t}\Rightarrow a=\frac{-2-10}{4}\Rightarrow a=\frac{-12}{4}=-3{ms}^{-2}$
From second law of motion, $F=m\times a\Rightarrow F=1\times (-3)=-3\text{Newtons}$
Here the minus sign indicates that the direction of force was opposite to the direction of initial velocity. Also, the magnitude of force was 3N (magnitude is the absolute value).