The correct option is A 1 N
Given,
Mass of body m=500 g=0.5 kg
Distance, S=4 m
Initial velocity, u=0 ms−1
Time taken, t=2 s.
Let force applied be F, acceleration be a and final velocity be v.
Applying, the second equation of motion, S=ut+12at2 and substituting the values we get,
4=0×2+12a×22
⇒ a=2 ms−2.
From Newton's second law of motion ,
F=m×a=0.5×2=1 N.
Therefore, the force applied is 1 N.