Solved on Jan 24, 2024

A rubber band exerts a restoring force $F=-20x+0.3x^2$ (in N) when stretched $x$ (in m). The work done stretching it from $x=0$ to $x=5$ is: a) 190 J, b) 356 J, c) 238 J, d) 119 J, e) 309 J.

STEP 1

Assumptions
1. The force exerted by the rubber band when stretched is given by the equation $F=-20x + 0.3x^2$ , where $x$ is in meters and $F$ is in Newtons.
2. Work done, $W$ , is calculated by integrating the force over the distance it acts upon.
3. The rubber band is stretched from $x=0$ to $x=5$ meters.

STEP 2

The work done by a variable force in one dimension is given by the integral of the force with respect to distance. The formula for work done, $W$ , is:
$W = \int_{x_1}^{x_2} F \, dx$
where $x_1$ and $x_2$ are the initial and final positions, respectively, and $F$ is the force as a function of $x$ .

STEP 3

Substitute the given force function into the work integral.
$W = \int_{0}^{5} (-20x + 0.3x^2) \, dx$

STEP 4

Calculate the integral term by term.
$W = \int_{0}^{5} (-20x) \, dx + \int_{0}^{5} (0.3x^2) \, dx$

STEP 5

Integrate the first term $-20x$ with respect to $x$ .
$\int_{0}^{5} (-20x) \, dx = \left[-10x^2\right]_{0}^{5}$

STEP 6

Evaluate the first integral from the lower limit $0$ to the upper limit $5$ .
$\left[-10x^2\right]_{0}^{5} = -10(5)^2 + 10(0)^2$

STEP 7

Calculate the result of the first integral.
$-10(5)^2 + 10(0)^2 = -10(25) = -250$

STEP 8

Integrate the second term $0.3x^2$ with respect to $x$ .
$\int_{0}^{5} (0.3x^2) \, dx = \left[0.1x^3\right]_{0}^{5}$

STEP 9

Evaluate the second integral from the lower limit $0$ to the upper limit $5$ .
$\left[0.1x^3\right]_{0}^{5} = 0.1(5)^3 - 0.1(0)^3$

STEP 10

Calculate the result of the second integral.
$0.1(5)^3 - 0.1(0)^3 = 0.1(125) = 12.5$

STEP 11

Add the results of the two integrals to find the total work done.
$W = -250 + 12.5$

STEP 12

Calculate the total work done, $W$ .
$W = -250 + 12.5 = -237.5$

STEP 13

Since work is a scalar quantity and cannot be negative in this context (as the force is in the direction of the displacement), we take the absolute value to find the magnitude of the work done.
$W = | -237.5 | = 237.5 \text{ J}$
The work done in stretching the rubber band from $x=0$ to $x=5$ is $237.5 \text{ J}$ , which corresponds to option c. $238 \text{ J}$ (rounded to the nearest whole number).

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A rubber band exerts a restoring force $F=-20x+0.3x^2$ (in N) when stretched $x$ (in m). The work done stretching it from $x=0$ to $x=5$ is: a) 190 J, b) 356 J, c) 238 J, d) 119 J, e) 309 J.

STEP 1

STEP 2

The work done by a variable force in one dimension is given by the integral of the force with respect to distance. The formula for work done, $W$ , is:
$W = \int_{x_1}^{x_2} F \, dx$
where $x_1$ and $x_2$ are the initial and final positions, respectively, and $F$ is the force as a function of $x$ .

STEP 3

Substitute the given force function into the work integral.
$W = \int_{0}^{5} (-20x + 0.3x^2) \, dx$

STEP 4

Calculate the integral term by term.
$W = \int_{0}^{5} (-20x) \, dx + \int_{0}^{5} (0.3x^2) \, dx$

STEP 5

Integrate the first term $-20x$ with respect to $x$ .
$\int_{0}^{5} (-20x) \, dx = \left[-10x^2\right]_{0}^{5}$

STEP 6

Evaluate the first integral from the lower limit $0$ to the upper limit $5$ .
$\left[-10x^2\right]_{0}^{5} = -10(5)^2 + 10(0)^2$

STEP 7

Calculate the result of the first integral.
$-10(5)^2 + 10(0)^2 = -10(25) = -250$

STEP 8

Integrate the second term $0.3x^2$ with respect to $x$ .
$\int_{0}^{5} (0.3x^2) \, dx = \left[0.1x^3\right]_{0}^{5}$

STEP 9

Evaluate the second integral from the lower limit $0$ to the upper limit $5$ .
$\left[0.1x^3\right]_{0}^{5} = 0.1(5)^3 - 0.1(0)^3$

STEP 10

Calculate the result of the second integral.
$0.1(5)^3 - 0.1(0)^3 = 0.1(125) = 12.5$

STEP 11

Add the results of the two integrals to find the total work done.
$W = -250 + 12.5$

STEP 12

Calculate the total work done, $W$ .
$W = -250 + 12.5 = -237.5$

STEP 13

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A rubber band exerts a restoring force F=−20x+0.3x2F=-20x+0.3x^2F=−20x+0.3x2 (in N) when stretched xxx (in m). The work done stretching it from x=0x=0x=0 to x=5x=5x=5 is: a) 190 J, b) 356 J, c) 238 J, d) 119 J, e) 309 J.

STEP 1

STEP 2

STEP 3

Substitute the given force function into the work integral.W=∫05(−20x+0.3x2) dxW = \int_{0}^{5} (-20x + 0.3x^2) \, dxW=∫05​(−20x+0.3x2)dx

STEP 4

Calculate the integral term by term.W=∫05(−20x) dx+∫05(0.3x2) dxW = \int_{0}^{5} (-20x) \, dx + \int_{0}^{5} (0.3x^2) \, dxW=∫05​(−20x)dx+∫05​(0.3x2)dx

STEP 5

Integrate the first term −20x-20x−20x with respect to xxx.∫05(−20x) dx=[−10x2]05\int_{0}^{5} (-20x) \, dx = \left[-10x^2\right]_{0}^{5}∫05​(−20x)dx=[−10x2]05​

STEP 6

Evaluate the first integral from the lower limit 000 to the upper limit 555.[−10x2]05=−10(5)2+10(0)2\left[-10x^2\right]_{0}^{5} = -10(5)^2 + 10(0)^2[−10x2]05​=−10(5)2+10(0)2

STEP 7

Calculate the result of the first integral.−10(5)2+10(0)2=−10(25)=−250-10(5)^2 + 10(0)^2 = -10(25) = -250−10(5)2+10(0)2=−10(25)=−250

STEP 8

Integrate the second term 0.3x20.3x^20.3x2 with respect to xxx.∫05(0.3x2) dx=[0.1x3]05\int_{0}^{5} (0.3x^2) \, dx = \left[0.1x^3\right]_{0}^{5}∫05​(0.3x2)dx=[0.1x3]05​

STEP 9

Evaluate the second integral from the lower limit 000 to the upper limit 555.[0.1x3]05=0.1(5)3−0.1(0)3\left[0.1x^3\right]_{0}^{5} = 0.1(5)^3 - 0.1(0)^3[0.1x3]05​=0.1(5)3−0.1(0)3

STEP 10

Calculate the result of the second integral.0.1(5)3−0.1(0)3=0.1(125)=12.50.1(5)^3 - 0.1(0)^3 = 0.1(125) = 12.50.1(5)3−0.1(0)3=0.1(125)=12.5

STEP 11

Add the results of the two integrals to find the total work done.W=−250+12.5W = -250 + 12.5W=−250+12.5

STEP 12

Calculate the total work done, WWW.W=−250+12.5=−237.5W = -250 + 12.5 = -237.5W=−250+12.5=−237.5

STEP 13

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A rubber band exerts a restoring force $F=-20x+0.3x^2$ (in N) when stretched $x$ (in m). The work done stretching it from $x=0$ to $x=5$ is: a) 190 J, b) 356 J, c) 238 J, d) 119 J, e) 309 J.

Substitute the given force function into the work integral.
$W = \int_{0}^{5} (-20x + 0.3x^2) \, dx$

Calculate the integral term by term.
$W = \int_{0}^{5} (-20x) \, dx + \int_{0}^{5} (0.3x^2) \, dx$

Integrate the first term $-20x$ with respect to $x$ .
$\int_{0}^{5} (-20x) \, dx = \left[-10x^2\right]_{0}^{5}$

Evaluate the first integral from the lower limit $0$ to the upper limit $5$ .
$\left[-10x^2\right]_{0}^{5} = -10(5)^2 + 10(0)^2$

Calculate the result of the first integral.
$-10(5)^2 + 10(0)^2 = -10(25) = -250$

Integrate the second term $0.3x^2$ with respect to $x$ .
$\int_{0}^{5} (0.3x^2) \, dx = \left[0.1x^3\right]_{0}^{5}$

Evaluate the second integral from the lower limit $0$ to the upper limit $5$ .
$\left[0.1x^3\right]_{0}^{5} = 0.1(5)^3 - 0.1(0)^3$

Calculate the result of the second integral.
$0.1(5)^3 - 0.1(0)^3 = 0.1(125) = 12.5$

Add the results of the two integrals to find the total work done.
$W = -250 + 12.5$

Calculate the total work done, $W$ .
$W = -250 + 12.5 = -237.5$