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adelina 88 [10]

3 years ago

8

A runaway train with a mass of 6,251 kg is approaching the end of the line at a speed of 27 m/s. The end of the line has a massi

ve spring with a spring constant of 48,979 Nm. How far will the spring be compressed (in meters) when the train comes to rest

1 answer:

stiv31 [10]3 years ago

7 0

Answer:

The compression in the spring is 9.64 meters.

Explanation:

Given that,

Mass of the train, m = 6251 kg

Speed of the train, v = 27 m/s

Spring constant of the spring, k = 48979 N/m

We need to find the compression in the spring when it comes to rest. It is a case of conservation of energy. The kinetic energy of the train is converted to elastic potential energy. So,

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

x is compression in spring

$mv^2=kx^2\\\\x=\sqrt{\dfrac{mv^2}{k}} \\\\x=\sqrt{\dfrac{6251\times (27)^2}{48979 }}\\\\x=9.64\ m$

So, the compression in the spring is 9.64 meters.

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12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

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h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

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$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

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$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

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Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

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From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

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Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

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$PE_1$ is the potential energy at the top of the 1st hill

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From the previous questions, we know that

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And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

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