Question

A body of mass $4kg$ is accelerated upon by a constant force, travels a distance of $5m$ in the first second and a distance of $2m$ in the third second. The force acting on the body is

• A. $2N$
• B. $4N$
• C. $6N$
• D. $8N$

Answer 1

Answer: (C)

$6N$

let the acceleration be a, so

$5=ut+\frac{a{t}^2}{2}=u+a/2$

$S_{3-2}=u(t_3-t_2)+\frac{a(t_3^2-t_2^2)}{2}=u+2.5a=2$

$a=-1.5$, So, $F=ma=-6N$