A spring has a natural length of 10in. An 800lb force stretches the spring to 14in.
A. Find the force constant
B. How much work is done in stretching the spring from 10 to 12 inches?
C. How far beyond its natural length will a 1600lb force stretch the spring.
A. To find the force constant (k) of the spring, we can use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement from its natural length.
Hooke's Law equation: F = -kx
Where:
F is the force acting on the spring
k is the force constant
x is the displacement from the natural length
In this case, we know that the natural length (x) of the spring is 10in and the force (F) is 800lb when the spring is stretched to 14in. Plugging in these values into Hooke's Law:
800 = -k * (14 - 10)
Simplifying the equation:
800 = -4k
Dividing both sides of the equation by -4:
k = -800 / -4
k = 200 lb/in
Therefore, the force constant (k) of the spring is 200 lb/in.
B. To calculate the work done in stretching the spring from 10 to 12 inches, we can use the formula for calculating work done by a spring:
Work (W) = (1/2) * k * (x2^2 - x1^2)
Where:
W is the work done
k is the force constant
x2 is the final displacement
x1 is the initial displacement
In this case, the initial displacement (x1) is 10in and the final displacement (x2) is 12in. Plugging in these values into the formula:
W = (1/2) * 200 * (12^2 - 10^2)
W = (1/2) * 200 * (144 - 100)
W = (1/2) * 200 * 44
W = 4400 lb-in
Therefore, the work done in stretching the spring from 10 to 12 inches is 4400 lb-in.
C. To determine how far beyond its natural length a 1600lb force will stretch the spring, we can rearrange Hooke's Law:
F = -kx
Solving for x:
x = -F / k
Plugging in the values F = 1600lb and k = 200 lb/in:
x = -1600 / 200
x = -8in
The negative sign indicates that the spring will stretch in the opposite direction. So, a 1600lb force applied to the spring will stretch it 8 inches beyond its natural length.