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A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving horizontally withconstant acceleration 4 m/s2. A bead can slide on the rod, and friction coefficient between them is 1/2.If the bead is released from rest at the top of the rod, find the time when it will reach at the bottom. Solution) If mass of bead is m;mg-µN=maNow, N=pseudo force due to acceleration of train, which is equal to 4m Newton. => mg-µ(4m)=ma=> g-4µ=a=> a=8ms-2 Using, second equation of motion:1=0.5(8)t2T=0.5sec
.how can we prove that sn=u+a(1/2)
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