A block of weight 3.9 {N} is launched up a 30(degree) inclined plane of length 2.20 {m} by a spring with spring constant 2.05 {kN/m} and maximum compression 0.10 {m}. The coefficient of kinetic friction is 0.50.
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Energy stored in spring Es = ½ke² (J)
Es = ½(2.05^3 N/m)(0.10m)² = 10.25 J
Energy dissipated by friction Ef = F*d
Ef = (μ.mg)*2.20m
Ef = (0.50*3.90)(2.2) = 4.29 (J)
Energy transferred to Grav.PE Eg = mg.h
Eg = 3.9*(2.20*sin30) = 4.29 (J)
Kinetic Energy Ek = Es - Ef - Eg
Ek = 10.25 - 4.29 - 4.29 .. .. .. ►Ek = 1.67 J
Es = ½(2.05^3 N/m)(0.10m)² = 10.25 J
Energy dissipated by friction Ef = F*d
Ef = (μ.mg)*2.20m
Ef = (0.50*3.90)(2.2) = 4.29 (J)
Energy transferred to Grav.PE Eg = mg.h
Eg = 3.9*(2.20*sin30) = 4.29 (J)
Kinetic Energy Ek = Es - Ef - Eg
Ek = 10.25 - 4.29 - 4.29 .. .. .. ►Ek = 1.67 J
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u need the equation