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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Work, Power and Energy

BITSAT 2013 · Physics · Work, Power and Energy

A 3.628 kg freight car moving along a horizontal rail road spur...

A 3.628kg3.628 \mathrm{ kg} freight car moving along a horizontal rail road spur track at 7.2km/7.2 \mathrm{ km} / hour strikes a bumper whose coil springs experiences a maximum compression of 30cm30 \mathrm{ cm} in stopping the car. The elastic potential energy of the springs at the instant when they are compressed 15cm15 \mathrm{ cm} is

Options

  1. A12.1×104J12.1 \times 10^{4} \mathrm{ J}
  2. B121×104J121 \times 10^{4} \mathrm{ J}
  3. C1.21×104J1.21 \times 10^{4} \mathrm{ J}
  4. D1.21×106J1.21 \times 10^{6} \mathrm{ J}

Correct answer

C. 1.21×104J1.21 \times 10^{4} \mathrm{ J}

Step-by-step solution

$\begin{aligned}7.2 \mathrm{ km}/ \mathrm{ram}&=\frac{7.2 \times \frac{5}{18} \mathrm{ m} / \mathrm{s}}{7 / 2^{2}}\\ &=\frac{72 \times 5}{10 \times 18}=2 \mathrm{ m}/ \mathrm{s} \end{aligned} firt case EC (\frac{1}{2}k x^{2}=\frac{1}{2}m v^{2}) (\Rightarrow \frac{1}{7}k\left(30 \times10^{-2}\right)^{2}=\frac{1}{2} \times3.628 \times(2)^{2}) (k=\frac{4 \times 3.628}{9 \times 10^{-2}})  EC 12kx2=12mv2⇒12k(30×10−2)2=12×3.628×(2)2k=4×3.6289×10−2\begin{aligned}\text{ EC }& \frac{1}{2} k x^{2}=\frac{1}{2} m v^{2} \\ \Rightarrow & \frac{1}{2} k\left(30 \times10^{-2}\right)^{2}=\frac{1}{2} \times3.628 \times(2)^{2} \\ k &=\frac{4 \times 3.628}{9 \times 10^{-2}} \end{aligned}energypotentialenrgywhenconpressedat(15cm)energy potential enrgy when conpressed at (15 \mathrm{ cm})⇒12k(15×10−2)2=12×4×3.6289×10−2×225×10−4=1639×10−29=182.206×10−2\begin{aligned}\Rightarrow& \frac{1}{2} k\left(15 \times10^{-2}\right)^{2} \\=& \frac{1}{2} \times \frac{4 \times 3.628}{9 \times 10^{-2}} \times225 \times10^{-4} \\=& \frac{1639 \times 10^{-2}}{9}\\=& 182.206 \times10^{-2} \end{aligned}$)⇒12k(15×102=12×α×3.6289×10−2×225×10−4=1639×10−29=182.206×10−2\begin{aligned}&\Rightarrow \frac{1}{2} k\left(15 \times10^{2}\right. \\ &=\frac{1}{2} \times \frac{\alpha \times 3.628}{9 \times 10^{-2}} \times225 \times10^{-4} \\ &=\frac{1639 \times 10^{-2}}{9}\\ &=182.206 \times10^{-2} \end{aligned}$$)
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