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Please show the steps to the answer? I would appreciate learning how to do this right.
Please show the steps to the answer? I would appreciate learning how to do this right.
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Calculate TOTAL Mechanical Energy (KE + GPE) of cart on each peak (K & L).
at K: KE = 1/2mV² = (0.5)(315)(17.3)² = 47,138 J
at K: GPE = mgh = (315)(9.81)(44) = 135,967 J
at K: Total ME = KE + GPE = 47,138 + 135,967 = 183,105 J
at L: KE = 1/2mV³ = (0.5)(315)(12.6)² = 25,005 J
at L: GPE = mgh = (315)(9.81)(30) = 92,705 J
at L: Total ME = KE + GPE = 117,710 J
Conservation of Energy would mean total ME's would be equal, but since they are not, there is a leak of energy in the system caused by a non-conserving force (friction).
Difference in Total ME= 183,105-117,710 = 65,396 J = lost energy due to friction ANS
Due to significance of data this ANS should be reported as = 65,400 J
at K: KE = 1/2mV² = (0.5)(315)(17.3)² = 47,138 J
at K: GPE = mgh = (315)(9.81)(44) = 135,967 J
at K: Total ME = KE + GPE = 47,138 + 135,967 = 183,105 J
at L: KE = 1/2mV³ = (0.5)(315)(12.6)² = 25,005 J
at L: GPE = mgh = (315)(9.81)(30) = 92,705 J
at L: Total ME = KE + GPE = 117,710 J
Conservation of Energy would mean total ME's would be equal, but since they are not, there is a leak of energy in the system caused by a non-conserving force (friction).
Difference in Total ME= 183,105-117,710 = 65,396 J = lost energy due to friction ANS
Due to significance of data this ANS should be reported as = 65,400 J
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Apply conservation of energy ..
Total initial (KE+PE) = final (KE+PE) + Energy transferred to heat by friction
½m.v1² + mg.h1 = ½m.v2² + mg.h2 + E
½m(v1² - v2²) + mg(h1 - h2) = E
½ 315(17.3² - 12.6²) + {315 x 9.80 x (44 - 30)} = E
22133.5 + 43218 .. .. ►E = 65351.50J .. (6.54^4 J)
Total initial (KE+PE) = final (KE+PE) + Energy transferred to heat by friction
½m.v1² + mg.h1 = ½m.v2² + mg.h2 + E
½m(v1² - v2²) + mg(h1 - h2) = E
½ 315(17.3² - 12.6²) + {315 x 9.80 x (44 - 30)} = E
22133.5 + 43218 .. .. ►E = 65351.50J .. (6.54^4 J)