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The barricade at the end of a subway line has a large spring designed to compres

ID: 1470768 • Letter: T

Question

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 2.05 times 10^5 kg train moving at 0.550 m/s. What is the force constant of the spring? What speed would the train be going if it only compressed the spring 0.740 m? How does the kinetic energy of the train before impact with the spring compare to the final potential energy of the spring when the train is stopped? m/s What force does the spring exert when compressed 0.740 m? (Enter the magnitude only.) How can you determine the magnitude and direction of the force exerted by a compressed spring? N

Explanation / Answer

We would use k=mg/x when gravity is involved, for example holding a spring in the air with a spring attached.

instead We could use conservation of energy.

Ei=Ef
1/2mv^2 = 1/2kx^2
1/2(2.05*10^5)(.55^2) = 1/2k(2^2)
31006.25 = 1/2k(2^2)

31006.25= 2k

k = 15503.125N/m

b) use the total energy to find the speed

E = 1/2kx^2 + 1/2mv^2
31006.25 = 1/2(15503.125)(.74^2) + 1/2(2.05*10^5)v^2

31006.25= 4244.75 + 102500v^2

31006.25 - 4244.75 = 102500v^2

or v=sqrt31006.25 - 4244.75/ 102500
v = 0.51
=>therefore the speed is 0.51m/s

c)useing F = kx
F = (15503.125)(.74)
F = 11472.31N

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