Question #128093

A 500-ton train and collides with a buffer stop while traveling at 2 m/s. A buffer stop is a safety device and is designed to stop trains in case of brake failures. The mass of the buffer stop is 1.5 tons. It acts as a spring with a spring constant k = 1 x 106 N/m.

a. Estimate the mean deceleration, in [m/s2], experienced by the train as it is halted.

b. Estimate the distance, in [m], by which the buffer stop spring is compressed.

a. Estimate the mean deceleration, in [m/s2], experienced by the train as it is halted.

b. Estimate the distance, in [m], by which the buffer stop spring is compressed.

Expert's answer

The kinetic energy of the train got stored as elastic potential energy of the spring as a result of stopping it. So, if the spring gets compressed by a distance of x and it's force constant is k, its elastic potential energy stored will be "0.5kx^{2}". And if the train of mass m was moving with a speed of v, then its kinetic energy was "0.5mv^{2}" . So, equating both, we get:

"0.5mv^{2} = 0.5kx^{2}"

"0.5\\times500\\times10^{3}\\times2^{2}=0.5\\times1\\times10^{6}\\times x^{2}"

x=1.41m

Answer b) The distance, in [m], by which the buffer stop spring is compressed = 1.41m

Then we can calculate the deceleration of the train:

"v^{2}=u^{2}+2as" , where v - final speed, u - initial speed, a - acceleration, s - distance.

"0^{2}=2^{2}+2\\times a\\times 1.41"

a=-"1.41\\frac{m}{s^{2}}"

Answer a) Deceleration "1.41\\frac{m}{s^{2}}"

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