1. (a) Noting that the vertical displacement is 10.0 m – 1.50 m = 8.50 m downward (same direction as GFg), Eq. 7-12 yields 2cos(2.00 kg)(9.80 m/s )(8.50 m)cos0167 J.gWmgdφ==°=(b) One approach (which is fairly trivial) is to use Eq. 8-1, but we feel it is instructive to instead calculate this as ΔUwhere U= mgy(with upwards understood to be the +ydirection). The result is 2()(2.00 kg)(9.80 m/s )(1.50 m 10.0 m)167 J.fiUmgy yΔ=− =−=−(c) In part (b) we used the fact that Ui= mgyi=196 J. (d) In part (b), we also used the fact
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