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The work done by stretching a certain spring increases by
0.13 x joules per centimeter (where
x is the displacement, in centimeters, beyond the spring's natural length).
How much work (in Joules) must be done in order to stretch the spring from
x=4 to
x=10 ?
Choose 1 answer:
(A) 5.46
(B) 7.54
(C)
10.92
(D)
15.08

The work done by stretching a certain spring increases by $0.13 x$ joules per centimeter (where $x$ is the displacement, in centimeters, beyond the spring's natural length). $\newline$ How much work (in Joules) must be done in order to stretch the spring from $x=4$ to $x=10$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $5$ . $46$ $\newline$ (B) $7$ . $54$ $\newline$ (C) $\mathbf{1 0 . 9 2}$ $\newline$ (D) $\mathbf{1 5 . 0 8}$

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Area of quadrilaterals and triangles: word problems

Full solution

Q. The work done by stretching a certain spring increases by

0.13 x

joules per centimeter (where

x

is the displacement, in centimeters, beyond the spring's natural length).

\newline

How much work (in Joules) must be done in order to stretch the spring from

x=4

to

x=10

?

\newline

Choose

1

answer:

\newline

(A)

5

.

46

\newline

(B)

7

.

54

\newline

(C)

\mathbf{1 0 . 9 2}

\newline

(D)

\mathbf{1 5 . 0 8}

Understand the problem: Understand the problem. $\newline$ We are given a spring where the work done to stretch it increases by $0.13$ times the displacement in joules per centimeter. We need to calculate the total work done when the spring is stretched from $4\,\text{cm}$ to $10\,\text{cm}$ .
Set up the integral: Set up the integral to calculate the work done. $\newline$ The work done on the spring from $x=4$ to $x=10$ can be calculated by integrating the work done per centimeter over the displacement. The integral will be from $4$ to $10$ of $0.13x \, dx$ .
Calculate the integral: Calculate the integral. $\newline$ The integral of $0.13x$ with respect to $x$ is $0.13 \times (x^2/2)$ . We need to evaluate this from $x=4$ to $x=10$ .
Evaluate the integral: Evaluate the integral from $x=4$ to $x=10$ . Plugging in the values we get: $0.13 \times (10^2/2) - 0.13 \times (4^2/2)$ = $0.13 \times (100/2) - 0.13 \times (16/2)$ = $0.13 \times 50 - 0.13 \times 8$ = $6.5 - 1.04$ = $5.46$ joules

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