A.) How much work must be done in order to compress a spring with an elastic constant of 400 lb / in and 1.75 inches?
*Note: 1.75" is the compression & 400#/in is the k constant.
B.) If 50 ft lbs of work is done on the same spring (in part a) in order to compress it, by how much is the spring shortened?
(a)
W=kx^2 (watch units)
(b)
use same equation, with W=50 ft-lbs = 600 ft-in.
solve for x.
Correction to equation
(a)
W=kx^2 ÷2 (watch units)
(b)
use same equation, with W=50 ft-lbs = 600 ft-in.
solve for x.
So "MathMate" you're saying that the equation I should use is the following:
A) 1/2 * 400 #/in * 1.75"^2 = 612.5 lbs
B) 1/2 * 600 #/in * 1.75"^2 = 918.75 lbs
There is no subtraction between the two? for part b, I need the length of the spring not lbs...
To solve these questions, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. Mathematically, this can be written as F = kx, where F is the force applied on the spring, k is the spring constant, and x is the displacement (compression or elongation) of the spring.
Let's start by answering the first question:
A.) How much work must be done in order to compress a spring with an elastic constant of 400 lb/in and a compression of 1.75 inches?
To calculate the amount of work done, we need to find the force applied and the distance over which this force is applied.
1. First, we find the force applied (F) using Hooke's Law.
F = k * x, where F is the force and x is the displacement.
F = 400 lb/in * 1.75 in = 700 lb
2. Next, we find the distance over which the force is applied (d).
Since the question doesn't provide this information, let's assume the spring is compressed to its maximum potential. In that case, the distance over which the force is applied is equal to the displacement of the spring, which is 1.75 inches.
Now we can calculate the work done (W) using the formula:
W = F * d, where W is the work done, F is the force, and d is the distance.
W = 700 lb * 1.75 in = 1225 lb-in or in-lb (depending on how the force is applied)
Therefore, the work done in order to compress the spring by 1.75 inches is 1225 lb-in or in-lb (depending on force direction).
Moving on to the second question:
B.) If 50 ft-lbs of work is done on the same spring (from part A) in order to compress it, by how much is the spring shortened?
To find the amount of compression, we need to rearrange the work formula:
W = F * d
d = W / F
1. Convert 50 ft-lbs to lb-in.
Since 1 ft = 12 in, 1 ft-lb = 12 in-lb.
50 ft-lb = 50 * 12 lb-in = 600 lb-in
2. Now, using the formula, we can find the compression (d).
d = W / F = 600 lb-in / 700 lb = 0.857 inches (approximately)
Therefore, the spring is shortened by approximately 0.857 inches when 50 ft-lbs of work is done on it.