The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Force (F) = 3 pounds
- Extension (e) = 0.6 feet
- Work done (Wd) =?
<h3>How to determine the spring constant</h3>
- Force (F) = 3 pounds
- Extension (e) = 0.6 feet
- Spring constant (K) =?
F = Ke
Divide both sides by e
K = F/ e
K = 3 / 0.6
K = 5 pound/foot
Thus, the spring constant of the spring is 5 pound/foot
<h3>How to determine the work done</h3>
- Spring constant (K) = 5 pound/foot
- Extention (e) = 0.7 feet
- Work done (Wd) =?
Wd = ½Ke²
Wd = ½ × 5 × 0.7²
Wd = 2.5 × 0.49
Wd = 1.23 foot-pound
Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound
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Answer:
12x^2+9x^2-25 (quadratic equation)
a=12, b=9, c=-25
put this in quadratic formula
Step-by-step explanation:
The answer is P= 80/19
hope this helps
Answer:
= 2(2x + 5)
Step-by-step explanation:
Given that:
= (7x + 5) - (3x - 5)
By simplifying
"-" sign before bracket will alter the inner signs
= 7x + 5 - 3x + 5
Adding like terms
= 4x + 10
By taking 2 common we get:
= 2(2x + 5)
i hope it will help you!
