Question

An object of mass 1 kg was moving at a speed of $10\frac{m}{s}$. A constant force acts for 4s on the object and the speed of object was $2\frac{m}{s}$ in opposite direction after the applied force was removed. The force acting on the object was:

• A. 3N
• B. -3N
• C. 30N
• D. -30N

Answer 1

Answer: (A), (D)

The correct options are
A

3N

D

-30N

Initial velocity u= $10\frac{m}{s}$
Final velocity v= $-2\frac{m}{s}$ (since this velocity is in opposite direction of the initial velocity which is taken as positive)
t = 4s
$a=\frac{v-u}{t}\Rightarrow a=\frac{-2-10}{4}\Rightarrow a=\frac{-12}{4}=-3\frac{m}{s^2}F=m\times a\Rightarrow F=1\times (-3)=-3\text{Newtons}$
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).