Math  /  Calculus

Question(1 point) A force of 1 pounds is required to hold a spring stretched 0.4 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 1 feet beyond its natural length? \square Preview My Answers Submit Answers
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Studdy Solution

STEP 1

1. The spring follows Hooke's Law, which states that the force required to stretch or compress a spring is proportional to the distance it is stretched or compressed.
2. The spring constant k k can be determined from the given force and displacement.
3. Work done on the spring is calculated using the integral of the force over the distance stretched.

STEP 2

1. Determine the spring constant k k .
2. Set up the integral to calculate the work done.
3. Evaluate the integral to find the work done.

STEP 3

Determine the spring constant k k using Hooke's Law:
F=kx F = kx
Given that a force of 1 1 pound is required to stretch the spring 0.4 0.4 feet, we have:
1=k×0.4 1 = k \times 0.4
Solve for k k :
k=10.4=2.5pounds per foot k = \frac{1}{0.4} = 2.5 \, \text{pounds per foot}

STEP 4

Set up the integral to calculate the work done in stretching the spring from its natural length to 1 1 foot beyond its natural length. The work done W W is given by:
W=01kxdx W = \int_{0}^{1} kx \, dx
Substitute k=2.5 k = 2.5 :
W=012.5xdx W = \int_{0}^{1} 2.5x \, dx

STEP 5

Evaluate the integral:
W=012.5xdx W = \int_{0}^{1} 2.5x \, dx
=2.5[x22]01 = 2.5 \left[ \frac{x^2}{2} \right]_{0}^{1}
=2.5(122022) = 2.5 \left( \frac{1^2}{2} - \frac{0^2}{2} \right)
=2.5×12 = 2.5 \times \frac{1}{2}
=1.25foot-pounds = 1.25 \, \text{foot-pounds}
The work done in stretching the spring is:
1.25foot-pounds \boxed{1.25 \, \text{foot-pounds}}

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