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?
*Information on part (a) is 400 #/in is the k constant and 1.75" is the compression.
50 ft lbs = 600 in lbs
equation is: 1/2 * 600" lb * 1.75"^2 = 918.75 lbs???
I'm looking for the length in this question, how do I get that?
To find out how much the spring is shortened, you can use Hooke's Law, which states that the force needed to compress or extend a spring is directly proportional to the change in length of the spring.
The formula for Hooke's Law is as follows:
F = k * x
Where:
F is the force applied on the spring,
k is the spring constant,
x is the change in length of the spring.
In this case, you are given the spring constant, k, as 400 #/in and the length compression, x, as 1.75 inches.
Now, let's calculate the force applied on the spring. As you mentioned, 50 ft-lbs is equivalent to 600 in-lbs.
So, if we substitute the values into the formula, we have:
600 in-lbs = 400 #/in * x
To solve for x, we need to divide both sides of the equation by 400 #/in:
x = 600 in-lbs / 400 #/in
This simplifies to:
x = 1.5 inches
Therefore, the spring is shortened by 1.5 inches.