Maximum stress σmax=4×108 N/m2
Radius of wire r=4×10−3 m
Mass of lift m=1600 kg
Maximum tension the wire can withstand is given by:
Tmax=σmax×πr2
Tmax=4×108×3.14×(4×10−3)2
Tmax=4×108×3.14×16×10−6=20096 N
When the lift accelerates upwards with maximum acceleration a, the tension in the wire is:
Tmax=m(g+a)
20096=1600(10+a)
10+a=160020096=12.56
a=12.56−10=2.56 m/s2
Answer: 2.56