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. 2 N
• B. 4 N
• C. 6 N
• D. 8 N

Answer 1

Answer: (C)

Distance travelled by the body in $n^{\text{th }}$ second is given by
$S_n=u+\frac{a}{2}(2n-1)$
$5=u+\frac{a}{2}(2\times 1-1)$
$5=u+\frac{a}{2}...(i)$
$2=u+\frac{a}{2}(2\times 3-1)$
$2=u+\frac{5}{2}a....(ii)$
Solving Eqs. (i) and (ii), we get
$a=-\frac{6}{4}m/s^2$
ie, body is decelerating
mass $=4kg$
and$F=m\times a=4\times \frac{6}{4}=6N$