Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AIIMS 2005
Exam
AIIMS
Year
2005
Subject
Physics
Chapter
Work, Power and Energy

AIIMS 2005 · Physics · Work, Power and Energy

A block of mass 10 ~kg is moving in x -direction with a constant...

A block of mass 10 kg10 \mathrm{~kg} is moving in xx-direction with a constant speed of 10 m/sec10 \mathrm{~m} / \mathrm{sec}. It is subjected to a retarding force F=−0.1xF=-0.1 x joule/m eter during its travel from x=20x=20 meter to x=30x=30 meter. Its final kinetic energy will be

Options

  1. A475 joule
  2. B450 joule
  3. C275 joule
  4. D250 joule.

Correct answer

A. 475 joule

Step-by-step solution

According to work energy theorem, W=ΔK⋅E.\RightarrowW=12mv22−12mv12⇒F\cdotdx=K⋅E\cdotf−12×10×100⇒F\cdotdx=K⋅E\cdotf−500⇒∫2030−0.1xdx=K⋅E\cdotf−500⇒−0.1[x22]2030=K⋅E\cdotf−500⇒−0.1[900−4002]=K⋅E\cdotf−500\begin{aligned}& W=\Delta \mathrm{K} \cdot \mathrm{E} . \quad \RightarrowW=\frac{1}{2}m v_2^2-\frac{1}{2}m v_1^2 \\ \Rightarrow & F \cdotd x=\mathrm{K} \cdot \mathrm{E} \cdotf-\frac{1}{2} \times10 \times100 \\ \Rightarrow & F \cdotd x=\mathrm{K} \cdot \mathrm{E} \cdotf-500 \\ \Rightarrow & \int_{20}^{30}-0.1 x d x=\mathrm{K} \cdot \mathrm{E} \cdotf-500 \\ \Rightarrow & -0.1\left[\frac{x^2}{2}\right]_{20}^{30}=\mathrm{K} \cdot \mathrm{E} \cdotf-500 \\ \Rightarrow & -0.1\left[\frac{900-400}{2}\right]=\mathrm{K} \cdot \mathrm{E} \cdotf-500 \end{aligned}\Rightarrow K⋅E⋅f=\mathrm{K}_{\cdot} \mathrm{E}_{\cdot f}=500-25=475 $\mathrm{~J}
Practice Work, Power and Energy on Quantrex Academy →

More from Work, Power and Energy

All Work, Power and Energy questions Full Work, Power and Energy list All AIIMS PYQs