Question

A block of mass 10kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = -0.1 x J/m during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be

• A. 475 J
• B. 450 J
• C. 275 J
• D. 250 J

Answer 1

The correct option is A

475 J

Step1: Given data

The mass of the block, $m=10kg$.

The block moves with a velocity of $v=10m/s$.

Retarding force $F=-0.1x$ J/m during its travel from $x=20mtox=30m$

Step2: Formula used

$$\begin{gathered} K.E.=\frac{1}{2}m{v}^2[whereK.E=kineticenergy,m=mass,v=velocity] \\ dW=\int fdx[whereW=workdone,f=force,x=displacement] \end{gathered}$$

Step3: Calculating Kinetic energy

Substituting the given values of m and v.

$$\begin{gathered} K.E.=\frac{1}{2}m{v}^2 \\ K.E.=\frac{1}{2}\times 10\times 10 \\ K.E.=500J \end{gathered}$$

Step4: Calculating work done

Force ($F=-0.1x$) and it is acted on the block from $x=20mtox=30m$. Therefore integrating the work done as per the above limits we get,

$$\begin{gathered} \int dW={\int }_{20}^{30}0.1x.dx \\ W=0.1{\int }_{20}^{30}x.dx \\ W=0.1{\left(\frac{x^2}{2}\right)}_{20}^{30} \\ W=0.05[{30}^2-{20}^2] \\ W=0.05[900-400] \\ W=25J \end{gathered}$$

Step5: Calculating final kinetic energy

This work done reduces the kinetic energy of the block as the force acting is in the opposite direction of motion of the block, therefore the final K.E is equal to

$$\begin{gathered} K.E.\left(final\right)=K.E.-W \\ K.E.\left(final\right)=500-25 \\ K.E.\left(final\right)=475J \end{gathered}$$

The final kinetic energy is 475 J.

Hence, option a is the correct answer.