A linear spring of stiffness k is designed to stop the 20-Mg railroad car traveling at 8 km/h within 400mm after impact. Find the smallest value of k that
will produce the desired result.
m = 20*10^6 g = 20*10^3 kg
v = 8000m/3600 s = 2.22 m/s
Ke = (1/2) m v^2 = 10*10^3 * 4.94
= 4.94 * 10^4 Joules
= (1/2) k x^2
x = .4 meter (wow!)
x^2 = .16
(1/2) x^2 = .08
so
k = 4.94 * 10^4/.08 =61.7 *10^4 N/m
= 617,000 N/m
Why did the railroad car go to therapy? Because it had some serious spring stiffness issues!
To find the value of k, let's first convert the velocity of the car to m/s. We have:
8 km/h = 8,000 m/3,600 s = 20/9 m/s
Now, let's look at the situation. The car needs to be stopped within 400 mm. We know that the force exerted by the spring is given by Hooke's Law:
F = k * x
Where F is the force, k is the spring stiffness, and x is the displacement. In this case, x would be 400 mm, which can be converted to 0.4 m.
Since we're dealing with stopping the car, we need to find the force required to bring the car to a halt. This force would be equal to the deceleration of the car times its mass:
F = m * a
We're given that the mass of the car is 20 Mg, which is 20,000 kg. To find the acceleration, we can use the kinematic equation:
v^2 = u^2 + 2*a*x
Since the car is coming to a stop (final velocity v = 0) from an initial velocity u = 20/9 m/s, we can rearrange the equation to solve for a:
a = -u^2 / (2x)
Plugging in the values, we have:
a = - (20/9)^2 / (2 * 0.4) ≈ -16.86 m/s^2
Now, we can find the force F:
F = m * a = 20,000 * -16.86 ≈ -337,200 N
Since the spring force should oppose the motion of the car, we take the absolute value of F:
F = 337,200 N
Now, let's find the spring stiffness k:
337,200 = k * 0.4
Solving for k, we get:
k = 337,200 / 0.4 = 843,000 N/m
So, the smallest value of k that will produce the desired result is approximately 843,000 N/m. However, remember that this answer assumes ideal conditions and neglects factors such as damping or non-linearity in the spring's behavior. So this spring stiffness value may be an approximation. But hey, at least the car won't be clowning around anymore!
To find the smallest value of k that will produce the desired result, we can use the principle of conservation of energy. The kinetic energy of the railroad car before impact is equal to the potential energy stored in the spring after the car comes to a stop.
Given:
Mass of the railroad car (m) = 20 Mg = 20,000 kg
Velocity of the railroad car (v) = 8 km/h = 8,000 m/3600 s = 2.22 m/s
Compression distance (x) = 400 mm = 0.4 m
1. Calculate the initial kinetic energy (KE_initial):
KE_initial = 0.5 * m * v^2
KE_initial = 0.5 * 20000 kg * (2.22 m/s)^2
2. Calculate the final potential energy (PE_final) stored in the spring when the car comes to a stop:
PE_final = 0.5 * k * x^2
3. Equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final
0.5 * 20000 kg * (2.22 m/s)^2 = 0.5 * k * (0.4 m)^2
4. Solve the equation for k:
k = (0.5 * 20000 kg * (2.22 m/s)^2) / (0.5 * (0.4 m)^2)
k = (0.5 * 20000 * 2.22^2) / 0.4^2
k = (0.5 * 20000 * 4.9284) / 0.16
k = 492.84
Thus, the smallest value of k that will produce the desired result is 492.84 N/m.
To find the smallest value of the stiffness constant k that will produce the desired result, we can use the principle of work and energy.
When the railroad car comes to a stop after impact, the work done by the spring is equal to the initial kinetic energy of the car. Let's calculate the initial kinetic energy of the car.
The mass of the railroad car is given as 20 Mg, which is equivalent to 20,000 kg. The velocity of the car is given as 8 km/h, which is equivalent to 8,000/3 m/s.
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2) * m * v^2
Plugging in the values:
KE = (1/2) * 20000 * (8000/3)^2
Next, since the work done by the spring is equal to the initial kinetic energy, we can set up the equation:
work done by spring = KE
The work done by the spring can be calculated as the product of the force exerted by the spring and the distance it moves. The force exerted by the spring is given by Hooke's Law: F = k * x, where k is the stiffness constant and x is the distance the spring moves. In this case, the spring moves a distance of 400 mm, which is equivalent to 0.4 m.
So, the work done by the spring is given by:
work done by spring = k * x
Setting the work done by the spring equal to the initial kinetic energy, we have:
k * 0.4 = (1/2) * 20000 * (8000/3)^2
Now, we can solve for the stiffness constant k:
k = [(1/2) * 20000 * (8000/3)^2] / 0.4
Evaluating the above expression will give us the smallest value of k that will produce the desired result.