Using Hooke's Law (F=kx) determine:
A) A weight of 5 pounds stretches a certain spring from its natural length of 10.2 inches to a length of 10.5 inches. Find K.
B) What weight will stretch the spring in part (A) to a length of 13.2 inches?
Thank you!
hmm. 5 = 0.3k
use that k to get
F = 13.2k
To determine the value of k using Hooke's Law (F = kx), you need to know the force applied (F) and the displacement (x) caused by that force.
For part A) given:
- Force (F) = 5 pounds
- Displacement (x) = (final length - initial length) = (10.5 inches - 10.2 inches)
Calculating the displacement:
x = 10.5 inches - 10.2 inches = 0.3 inches
Now, substituting the values into Hooke's Law, we have:
F = kx
5 pounds = k * 0.3 inches
To solve for k, divide both sides of the equation by 0.3 inches:
k = 5 pounds / 0.3 inches
Therefore, the value of k is equal to:
k = 5 pounds / 0.3 inches
Now, let's move on to part B) to find the weight required to stretch the spring to a length of 13.2 inches.
Given:
- Initial length (x1) = 10.2 inches
- Final length (x2) = 13.2 inches
- Value of k = calculated in part A) = (5 pounds / 0.3 inches)
To find the weight (F2) required, we'll rearrange the formula to solve for F:
F = kx
Substituting the values of k and x2:
F2 = (5 pounds / 0.3 inches) * 3 inches
Now, we can simplify the equation:
F2 = 5 pounds * 10
Therefore, the weight (F2) required to stretch the spring to a length of 13.2 inches is:
F2 = 50 pounds