Question

A body of mass $4kg$ is accelerated up 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

The correct option is C

$6N$

Step 1. Given data

Mass of the body, $m=4kg$

Distance travelled in first second, $S_1=5m$

Distance travelled in third second, $S_3=2m$

Step 2. Finding the value of acceleration, $a$ and the initial velocity, $u$

By using the formula of displacement of $n^{th}$second,

$S_n=u+\frac{a}{2}\left(2n-1\right)$

For the first second, $n=1$

$S_1=u+\frac{a}{2}\left(2\times 1-1\right)$

$5=u+\frac{a}{2}$ [$eq\left(1\right)$]

For the third second, $n=3$

$S_3=u+\frac{a}{2}\left(2\times 2-1\right)$

$2=u+\frac{5a}{2}$ [$eq\left(2\right)$]

Subtracting both the equations, we get

$a=-1.5m{s}^{-2}$

Substituting the value of $a$ in $eq\left(1\right)$, we get

$u=5.75m{s}^{-1}$

Step 3. Finding the force, $F$

By using the formula of force

$F=ma$

$F=4\times \left(-1.5\right)$

$F=-6N$ [The negative sign indicates that there is a deacceleration.]

Hence, the correct option is $\left(C\right)$.