Question

The distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 9 inches with 100 pounds attached, how far will it stretch with 90 pounds attached? Round to the nearest tenth of an inch.

A. 8.1 in.
B. 8.9 in.
C. 10 in.
D. 9.1 in.

Answer 1

Answer A - 8,1in

explanation
------------------------------------------------------------------------
! Inches ! Inches ! Pounds ! Pounds !
! 9 : 10 ! Sum ! ! sum !
-------------------------------------------------------------------------
! 0,9 ! 0,9*1 = 09 ! 10 ! 1*10 = 10 !
------------------------------------------------------------------------\
! 0,9 ! 0.9*2 = 1,8 ! 10 ! 2*10 = 2 0 !
! 0,9 ! ! 10 ! !
-----------------------------------------------------------------------
! 0,9 ! 0,9*3 = 2,7 ! 10 ! !
! 0,9 ! ! 10 ! 3*10 = 30 !
! 0,9 ! ! 10 ! !
-------------------------------------------------------------------------
for every pound falls 0.9 inches
derived unit of proportion :


`(9 inches)/(100 pounds) = (0,9 inches)/(10pounds) \\ \\ = (0,9 * 9 inches)/(10 *9 pounds) = (8,1 inches)/(90pounds)`

Answer 2

x2 = 8.1 in (Option A)

Explanation:

Given:-

- The extension of spring x1 = 9 in

- The weight hanged F1 = 100 lbs

Find:-

How far will it stretch with 90 pounds attached?

Solution:-

- The linear extension x of spring is proportional to the amount of force F applied to the spring.

F = k*x

Where, k = is the proportionality constant called stiffness of the spring.

- We will first determine the spring stiffness k using the given data as follows:

F1 = k*x1

100 = k*9

k = 100 / 9

- Now, we will determine the amount with which the spring stretches x2 when the hanged weight is changed to F2 = 90 lbs. Using the first relation we plug in the value of proportionality constant (stiffness) k.

F = (100 / 9)*x

F2 = (100 / 9)*x2

90 = (100 / 9)*x2

x2 = 810 / 100

x2 = 8.1 in

- The amount of spring stretches when the applied load of weight hanged is changed to F2 = 90 lbs is x2 = 8.1 in.