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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Work, Power and Energy

AIIMS 2019 · 25 May · Physics · Work, Power and Energy

Initially spring is in natural length and both the blocks are in...

Initially spring is in natural length and both the blocks are in rest condition. Then determine the maximum extension in the spring. K=20 N m−1K=20 \mathrm{~N} \mathrm{~m}^{-1} Initially spring is in natural length and both the blocks are in rest condition. Then determine the maximum extension

Options

  1. A203 cm\frac{20}{3} \mathrm{~cm}
  2. B103 cm\frac{10}{3} \mathrm{~cm}
  3. C403 cm\frac{40}{3} \mathrm{~cm}
  4. D193 cm\frac{19}{3} \mathrm{~cm}

Correct answer

A. 203 cm\frac{20}{3} \mathrm{~cm}

Step-by-step solution

a=Fm1+m2a=\frac{F}{m_1+m_2} By work - energy theorem (F−m1a)x1+(m2a)x2−12K(x1+x2)2=0m2×Fm1+m2(x1+x2)=K2(x1+x2)2(x1+x2)=m2×Fm1+m2×2k=1×11.5×220=115m=10015 cm=203 cm\begin{aligned} & \left(F-m_1 a\right) x_1+\left(m_2 a\right) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\ & m_2 \times \frac{F}{m_1+m_2}\left(x_1+x_2\right)=\frac{K}{2}\left(x_1+x_2\right)^2 \\ & \left(x_1+x_2\right)=m_2 \times \frac{F}{m_1+m_2} \times \frac{2}{k} \\ & =\frac{1 \times 1}{1.5} \times \frac{2}{20}=\frac{1}{15} m=\frac{100}{15} \mathrm{~cm}=\frac{20}{3} \mathrm{~cm}\end{aligned}
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