It takes a force of 21,714lb to compress a coil spring assembly on a New York City Transit Authority subway car from its height of 8in to its fully compressed height of 5in.
A. What is the assemblys force constant?
B. How much work does it take to compress the assembly the first half-inch? The second half-inch?
To find the force constant (k) of the coil spring assembly, we can use Hooke's Law, which states that the force needed to compress or extend a spring is directly proportional to the displacement.
Hooke's Law equation: F = k * x
where F is the force, k is the force constant, and x is the displacement.
Given:
Force (F) = 21,714 lb
Initial displacement (x1) = 8 in
Final displacement (x2) = 5 in
A. Calculating the force constant (k):
Using Hooke's Law, we can rearrange the equation:
k = F / x
Substituting the values:
k = 21,714 lb / (8 in - 5 in) = 21,714 lb / 3 in
k ≈ 7,238 lb/in
Therefore, the force constant of the coil spring assembly is approximately 7,238 lb/in.
B. Calculating the work done to compress the assembly:
The work done (W) is given by:
W = (1/2) * k * x^2
Where x is the displacement.
For the first half-inch (Δx = 0.5 in):
W1 = (1/2) * 7,238 lb/in * (0.5 in)^2
W1 = (1/2) * 7,238 lb/in * 0.25 in^2
W1 = 905 lb
Therefore, it takes approximately 905 lb of work to compress the assembly by the first half-inch.
For the second half-inch (Δx = 0.5 in):
W2 = (1/2) * 7,238 lb/in * (0.5 in)^2
W2 = (1/2) * 7,238 lb/in * 0.25 in^2
W2 = 905 lb
Similarly, it takes approximately 905 lb of work to compress the assembly by the second half-inch.
To find the force constant of the coil spring assembly, you can use Hooke's Law, which states that the force needed to compress or extend a spring is directly proportional to the displacement (change in length) of the spring. The formula for Hooke's Law is:
F = k * x
Where F is the force applied, k is the force constant (also known as the spring constant), and x is the displacement or change in length of the spring.
For part A, we can use the given information to find the force constant:
F = 21,714 lb
x = 8 in - 5 in = 3 in
Substituting these values into the equation, we have:
21,714 lb = k * 3 in
Dividing both sides by 3 in, we get:
k = 21,714 lb / 3 in
Therefore, the force constant of the coil spring assembly is approximately 7,238 lb/in.
For part B, we can calculate the work done to compress the assembly for the first half-inch and the second half-inch separately. The formula to calculate work is:
Work = Force * Distance
To calculate the work done to compress the assembly the first half-inch, we need to find the force and distance. The force can be calculated using Hooke's Law:
F = k * x = 7,238 lb/in * 0.5 in
Substituting the values, we get:
F = 3,619 lb
Now, we can calculate the work done:
Work = Force * Distance = 3,619 lb * 0.5 in
Work = 1,809.5 lb.in or 150.8 ft.lb (rounded to the nearest tenth)
To calculate the work done to compress the assembly the second half-inch, we repeat the same steps:
F = k * x = 7,238 lb/in * 0.5 in
F = 3,619 lb
Work = Force * Distance = 3,619 lb * 0.5 in
Work = 1,809.5 lb.in or 150.8 ft.lb (rounded to the nearest tenth)
Therefore, the work done to compress the assembly for both the first and second half-inch is approximately 1,809.5 lb.in or 150.8 ft.lb.
f = -k x
21714 = k (3)
W = (1/2) k x^2
do for x = 1/2 inch
then for 1 inch and subtract the half inch result