Solved on Jan 24, 2024

A rubber band exerts a restoring force F=20x+0.3x2F=-20x+0.3x^2 (in N) when stretched xx (in m). The work done stretching it from x=0x=0 to x=5x=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+0.3x2F=-20x + 0.3x^2, where xx is in meters and FF is in Newtons.
2. Work done, WW, is calculated by integrating the force over the distance it acts upon.
3. The rubber band is stretched from x=0x=0 to x=5x=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, WW, is:
W=x1x2FdxW = \int_{x_1}^{x_2} F \, dx
where x1x_1 and x2x_2 are the initial and final positions, respectively, and FF is the force as a function of xx.

STEP 3

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

STEP 5

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

STEP 6

Evaluate the first integral from the lower limit 00 to the upper limit 55.
[10x2]05=10(5)2+10(0)2\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)=250-10(5)^2 + 10(0)^2 = -10(25) = -250

STEP 8

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

STEP 9

Evaluate the second integral from the lower limit 00 to the upper limit 55.
[0.1x3]05=0.1(5)30.1(0)3\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)30.1(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 + 12.5

STEP 12

Calculate the total work done, WW.
W=250+12.5=237.5W = -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 JW = | -237.5 | = 237.5 \text{ J}
The work done in stretching the rubber band from x=0x=0 to x=5x=5 is 237.5 J237.5 \text{ J}, which corresponds to option c. 238 J238 \text{ J} (rounded to the nearest whole number).

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