Physics, 10.04.2020 05:07 ghari112345

A 250-kN railroad car A is traveling at 40 m/s while a 550-kN freight car B is traveling at 10 m/s (both cars heading to the right). If the spring constant mounted on car A is k = 70 MN/m, determine the maximum deformation undergone by the spring. Neglect friction.

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Answer from: keke6361

the maximum deformation undergone by the spring = 47.46 cm

Explanation:

Using conservation of momentum:

m_Av_A + m_Bv_B = (m_A+m_B )v

where:

$m_A = \frac{250\ kN}{g}$

v_A = 40

$m_B = \frac{550\ kN}{g}$

v_B =10

Then;

m_Av_A + m_Bv_B = (m_A+m_B )v

$\frac{250*10^3}{9.81}*40 + \frac{550*10^3}{9.81}*10 = (\frac{800*10^3}{9.81} )v$

$1580020.387 = 81549.43935 \ v$

$v = \frac{1580020.387}{81549.43935}$

v = 19.375 m/s

However ; using conservation of energy to determine the maximum deformation undergone by the spring ; we have:

$\frac{1}{2} [m_Av_A^2 +m_Bv_B^2] =\frac{1}{2}[(m_A+m_B)v^2 + kx^2]$

[m_Av_A^2 +m_Bv_B^2] =[(m_A+m_B)v^2 + kx^2]

$[\frac{250*10^3}{9.81}*40^2 + \frac{550*10^3}{9.81}*10^2] =[ (\frac{800*10^3}{9.81} )*19.375^2 + 70 *10^6 \ * x^2]$

x = 0.4746 m

x = 47.46 cm

Thus, the maximum deformation undergone by the spring = 47.46 cm

Answer from: rahulshahi7569

The maximum deformation undergone by the spring is 0.475 m

Explanation:

The conservation of momentum is equal:

$m_{A} v_{A} +m_{B} v_{B} =v(m_{A} +m_{B} )$

Replacing:

$\frac{250x10^{3}*40 }{9.8} +\frac{550x10^{3}*10 }{9.8} =v(\frac{(250+550)x10^{3} }{9.8}) \\v=19.375m/s$

The conservation of energy is:

$\frac{1}{2} (m_{A}v_{A}^{2}+m_{B}v_{B}^{2} )=\frac{1}{2} (v^{2}(m_{A} +m_{B} )+kx^{2} )\\\frac{250x10^{3}*40^{2} }{9.8} +\frac{550x10^{3}*10^{2} }{9.8} =19.375^{2} (\frac{800x10^{3} }{9.8} )+70x10^{6} x^{2} \\ x=0.475m$

Answer from: Quest

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Answer from: Quest

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A 250-kN railroad car A is traveling at 40 m/s while a 550-kN freight car B is traveling at 10 m/s (...

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