13. (II) The force on a particle, acting along the x axis, varies as shown in Fig. 6-38.Determine the work done by this force to move the particle along the x axis:(a) from x=0.0 to x=10.0m (b) from x=0.0 to x=15.0 FIGURE 6-38 Problem 13.

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13. (II) The force on a particle, acting along the x axis, varies as shown in Fig. 6-38.Determine the work done by this force to move the particle along the x axis:(a) from x=0.0 to x=10.0m (b) from x=0.0 to x=15.0 FIGURE 6-38 Problem 13.

Answer 1

To determine the work done by the force to move the particle along the x-axis, we need to calculate the area under the force vs. displacement graph between the specified limits of x. The work done by a force is given by the integral of the force with respect to displacement, which geometrically corresponds to the area under the force-displacement curve.Given the turning points and the shape of the graph, we can divide the graph into trapezoids and triangles to calculate the area.(a) From \(x = 0.0\) m to \(x = 10.0\) m:The graph increases linearly from (0,0) to (0,350), then remains constant at 350 N from (0,350) to (15,350). To find the work done from \(x = 0.0\) m to \(x = 10.0\) m, we only need to consider the area under the graph from \(x = 0.0\) m to \(x = 10.0\) m.This area is a right triangle with a base of 10.0 m (from 0 to 10 m) and a height of 350 N. The area \(A\) of a right triangle is given by:\[A = \frac{1}{2} \times \text{base} \times \text{height}\]Plugging in the values:\[A = \frac{1}{2} \times 10.0 \, \text{m} \times 350 \, \text{N}\]\[A = 5.0 \, \text{m} \times 350 \, \text{N}\]\[A = 1750 \, \text{J}\]So, the work done by the force to move the particle from \(x = 0.0\) m to \(x = 10.0\) m is 1750 J.(b) From \(x = 0.0\) m to \(x = 15.0\) m:For this part, we need to consider the entire area under the graph from \(x = 0.0\) m to \(x = 15.0\) m. This includes the triangle from part (a) and a rectangle from \(x = 10.0\) m to \(x = 15.0\) m.The area of the rectangle is given by:\[A_{\text{rectangle}} = \text{width} \times \text{height}\]\[A_{\text{rectangle}} = (15.0 \, \text{m} - 10.0 \, \text{m}) \times 350 \, \text{N}\]\[A_{\text{rectangle}} = 5.0 \, \text{m} \times 350 \, \text{N}\]\[A_{\text{rectangle}} = 1750 \, \text{J}\]Adding this to the work done from part (a):\[\text{Total work} = 1750 \, \text{J} + 1750 \, \text{J}\]\[\text{Total work} = 3500 \, \text{J}\]So, the work done by the force to move the particle from \(x = 0.0\) m to \(x = 15.0\) m is 3500 J.Final answers:(a) The work done from \(x = 0.0\) m to \(x = 10.0\) m is 1750 J.(b) The work done from \(x = 0.0\) m to \(x = 15.0\) m is 3500 J.

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13. (II) The force on a particle, acting along the x axis, varies as shown in Fig. 6-38.Determine the work done by this force to move the particle along the x axis:(a) from x=0.0 to x=10.0m (b) from x=0.0 to x=15.0 FIGURE 6-38 Problem 13.

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