Work, Power and EnergyMedium
Question
A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively. Then
Options
A.hA > hC ; KB > KC
B.hA > hC ; KC > KA
C.hA = hC ; KB = KC
D.hA < hC ; KB > KC
Solution
EA = mghA + KA
EB = KB
Ec = mghC + KC
Using conservation of energy
EA = EB = EC
KB > KC
KB > KA
Mg(hA - hC) + (KA - KC) = 0
⇒ hA - hC =
EB = KB
Ec = mghC + KC
Using conservation of energy
EA = EB = EC
KB > KC
KB > KA
Mg(hA - hC) + (KA - KC) = 0
⇒ hA - hC =
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