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Kinetic Energy) A body freely falls from a certain height ontothe ground in a time 't'. During the firstone third of the interval it gains a kineticenergy AK, and during the last one third ofthe interval, it gains a kinetic energy ΔΚ2.The ratio ΔΚ. AK, is1:5uin the mass of a boy and has |
Answer» speed gained from t = 0 , to t = t/3 => V = gt/3 so, K.E = 1/2*(m)*(gt/3)² = mg²t²/18 speed when t =2t/3 v = 0+g*2t/3 = 2gt/3 now speed gained from t=2/3 to t= t V = gt/3 +g2t/3 = gt so, K.E increased in last third = 1/2*m(g²t²) -1/2*m*(2gt/3)²= 1/2*m*g²t² -4/18m*g²*t² = 5/18mg²t². so, ∆k1:∆k2 = 1/18:5/18 = 1:5 option 4 |
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