Hooke's law states that the magnitude of force,F,required to stretch a spring x units beyond its natural length is directly proportional to x.If a force of 3 pounds stretches a spring from its natural length of 10 inches to a length of 10.7 inches,what force will stretch the spring to a length of 11.5 inches? Round your answer to the nearest hundredth.
A) $F = 5.52$
B) $F = 6.43$
C) $F = 5.70$
D) $F = 7.29$
E) $F = 6.14$

Question

Hooke's law states that the magnitude of force,F,required to stretch a spring x units beyond its natural length is directly proportional to x.If a force of 3 pounds stretches a spring from its natural length of 10 inches to a length of 10.7 inches,what force will stretch the spring to a length of 11.5 inches? Round your answer to the nearest hundredth.
A) $F = 5.52$
B) $F = 6.43$
C) $F = 5.70$
D) $F = 7.29$
E) $F = 6.14$ ​

Quizplus · Accepted Answer

Hooke's law is given by $F = kx$, where $F$ is the force applied, $x$ is the extension from the natural length, and $k$ is the spring constant. Given that a force of 3 pounds stretches the spring 0.7 inches (from 10 to 10.7 inches), we can find $k$ by rearranging the formula to $k = \frac{F}{x} = \frac{3}{0.7}$. To find the force required to stretch the spring to 11.5 inches, which is 1.5 inches beyond its natural length, we use the formula with the found value of $k$: $F = kx = \frac{3}{0.7} \times 1.5$, which gives approximately 6.43 pounds.

Hooke's Law States That the Magnitude of Force,F,required to Stretch F=5.52F = 5.52F=5.52

Hooke's Law States That the Magnitude of Force,F,required to Stretch $F = 5.52$