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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). (b) Where is the particle's turning point located? |
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Answer» Solution :KEY IDEA The turning point is where the force momentarity STOPS and then reverses the particle.s motion. That is , it iswhere the particle momentarily has `v=0` and thus `K = 0`. CALCULATIONS: Because K is the difference between `E_("mec")` and U, we want the point in Fig. 8-36a where the plot of U rises to MEET the horizontal line of `E_("mec")`, as shown in Fig. 8-36b, we can draw nested right triangles as shown and then write the proportionality of distances `(16-7.0)/(d) = (20-7.0)/(4.0 -1.0)` Which gives us `d=2.08m`. Thus, the turning point is at `x=4.0 m -d=1.9m`. |
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