10.16.Model: Assume an ideal spring that obeys Hooke’s law. Visualize:Solve: (a)The spring force on the 2.0 kg mass is sp.Fky=−ΔNotice that yΔis negative, so spFis positive. This force is equal to mg, because the 2.0 kg mass is at rest. We have .ky mg−Δ=Solving for k: 2(/)(2.0 kg)(9.8 m/s )/( 0.15 m( 0.10 m))392 N/mkmgy=−Δ =−−− −=The spring constant is 23.9 10 N/m.×(b) Again using :g−Δ =
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