Question

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

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

Answer 1

The Correct Option is D

Apply work-energy theorem. When a force acts upon a moving body, then the kinetic energy of the body increases and the increase is equal to the work done. This is work energy theorem. Work done $=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=K_{f}-K_{i}$ Another definition of work done is force $x$ displacement. $\therefore F d x=K_{f}-\frac{1}{2} m v_{i}^{2}$ where the subscripts $f$ and $i$ stand for final and initial. $F \cdot d x =K_{f}-\frac{1}{2} \times 10 \times(10)^{2}$ $\Rightarrow F \cdot d x=K_{f}-500$ $\Rightarrow \int\limits_{x=20}^{x=30}(-0.1) x d x =K_{f}-500$ Using the formula $\int x^{n} d x=\frac{x^{n+1}}{n+1}$, we have $-0.1\left[\frac{x^{2}}{2}\right]_{x=20}^{x=30} =K_{f}-500$ $-0.1\left[\frac{(30)^{2}}{2}-\frac{(20)^{2}}{2}\right] =K_{f}-500$ $\Rightarrow K_{f}-500 =-25$ $\Rightarrow K_{f}=500-25 =475\, J$