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A 2.00 kg particle moves along an x axis in onedimensional motion while a conservative force along that axis acts on it. The potential energy U(x) associated with the force is plotted in Fig. 8-36a. That is, if the particle were placed at any position between x=0 and x=7.00m, it would have the plotted value of U. At x=6.5 m, the particle has velocity vec(v)_(0) = (-4.00 m//s)hat(i). (c) Evaluate the force acting on the particle when it is in the region 1.9 m lt x lt 4.0 m. |
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Answer» Solution :KEY IDEA The force is given by Eq. 8-90 `(F(x)= - d U (x)//dx)`: The force is EQUAL to the negative of the slope on a graph of `U(x)`. Calculation: For the graph of Fig. 8-36b, we see that the RANGE `1.0 m LT x lt 4.0` m the force is `F= -(20J - 7.0 J)/(1.0 m - 4.0 m) = 4.3N`. Thus , the forcehas magnitude 4.3 N and is in the positive direction of the x axis . This result is consistent with the fact tha the initially leftward-moving PARTICLE is stopped by the force and then sent rightward. |
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