At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety device to stop the train so that it will not plow through the station if the engineer misjudges the stopping distance. While waiting, you wonder what would be the fastest train that the spring could stop at its full compression, 3.0 ft. To keep the passengers safe when the train stops, you assume a maximum stopping acceleration of g/2. You also guess that a train weighs half a million lbs. For purpose of getting an estimate, you decide to assume that all frictional force are negligible.
(a) What is the algebraic expression for the speed of the fastest train that could be stopped by the spring in terms of the maximum compression of the spring (L), the weight of the train (W), and the gravitational acceleration (g)? [Note: Don't enter an equation like "x=blah". Just enter the "blah" part. All letters are capital except for "g".]
(b) What is the numerical value of the speed of the fastest train that could be stopped by the spring (make sure to include units and put a space between the number and the units)?
rhg
To find the expression for the speed of the fastest train that could be stopped by the spring, we need to use the principle of conservation of mechanical energy.
(a) The expression for the velocity can be determined using the formula for the potential energy stored in a spring:
Potential energy = (1/2) * k * L^2
where k is the spring constant and L is the maximum compression of the spring.
However, we don't have the spring constant, k, directly given. Instead, we are given the weight of the train, W, and the maximum stopping acceleration, g/2. We can use these values to find the spring constant using the formula:
Weight of the train = k * L * (g/2)
Since we know the weight of the train is half a million pounds, we can substitute this value and solve for the spring constant:
500,000 lbs = k * 3.0 ft * (32.2 ft/s^2 / 2)
Simplifying this equation gives us the value of k:
k = (500,000 lbs) / (3.0 ft * (32.2 ft/s^2 / 2))
Now that we have the spring constant, we can use it to find the velocity expression.
The elastic potential energy stored in the spring is also equal to the kinetic energy of the moving train when it comes to a stop:
Potential energy = (1/2) * k * L^2 = (1/2) * m * v^2
where m is the mass of the train (converted from weight) and v is the velocity of the train.
To determine the expression for the velocity, we rearrange the equation:
v^2 = (2 * k * L^2) / m
Substituting the values of k, L, and m, we get:
v^2 = (2 * [(500,000 lbs) / (3.0 ft * (32.2 ft/s^2 / 2))] * (3.0 ft)^2) / (500,000 lbs / 32.2 ft/s^2)
Now, simplify and solve for v:
v^2 = 32.2 ft/s^2 * (2 * [(500,000 lbs) / (3.0 ft * (32.2 ft/s^2 / 2))] * (3.0 ft)^2) / (500,000 lbs)
Finally, taking the square root of v^2 will give us the expression for the velocity:
v = sqrt(32.2 ft/s^2 * (2 * [(500,000 lbs) / (3.0 ft * (32.2 ft/s^2 / 2))] * (3.0 ft)^2) / (500,000 lbs))
(b) To find the numerical value of the speed of the fastest train that could be stopped by the spring, we need to substitute the given values into the expression we derived in part (a).
Using the value of g = 32.2 ft/s^2, the weight of the train W = 500,000 lbs, and the maximum compression of the spring L = 3.0 ft, we can substitute these values into the expression:
v = sqrt(32.2 ft/s^2 * (2 * [(500,000 lbs) / (3.0 ft * (32.2 ft/s^2 / 2))] * (3.0 ft)^2) / (500,000 lbs))
Simplifying this expression will give us the numerical value of the speed of the fastest train that could be stopped by the spring in ft/s.