Question

An object is acted on by a retarding force of $10N$ and at a particular instant its kinetic energy is $6J$. The object will come to rest after it has travelled a distance of:

• A. $\frac{3}{5}$ m
• B. $\frac{5}{3}$ m
• C. $4$ m
• D. $16$ m

Answer 1

The correct option is A $\frac{3}{5}$ m

$\text{Step 1: Work-energy theorem}$

The object is coming to rest which means the final kinetic energy of the object is zero.

Applying the work-energy theorem,

$W=K_f-K_i$

$\Rightarrow W=0-6J$

$\Rightarrow W=-6J$ $....\left(1\right)$

$\text{Step 2: Distance travelled by the object}$

Let $|\overrightarrow{d}|=S$ be the distance travelled by the object.

We know work done by the force can be given as,

$W=\overrightarrow{F}.\overrightarrow{d}$

$W=FS\cos \theta$

Since $F$ is retarding force, the force and displacement would be in opposite direction and therefore $\theta ={180}^0$

$W=FS\cos {180}^0$

$\Rightarrow W=-10S$

From equation $\left(1\right)$$\Rightarrow -6J=-10S\Rightarrow S=\frac{3}{5}m$

Hence, Option A is correct.