An elevator weighing 4000 Ib attains an upward velocity of 4 m/sec in 3 sec withuniform acceleration. Find the apparent weight of a 40 kg man standing inside the elevator during its ascent
N−mg=ma→N=m(g+a)N-mg=ma\to N=m(g+a)N−mg=ma→N=m(g+a)
v=at→a=v/t=4/3≈1.33 (m/s2)v=at\to a=v/t=4/3\approx1.33\ (m/s^2)v=at→a=v/t=4/3≈1.33 (m/s2)
N=m(g+a)=40⋅(9.81+1.33)=445.73 (N)N=m(g+a)=40\cdot(9.81+1.33)=445.73\ (N)N=m(g+a)=40⋅(9.81+1.33)=445.73 (N) . Answer
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