Answer:
Given \[m=50g\frac{50}{1000}kg=\frac{1}{20}kg.\]. (a) Draw F.B.D in case of elevator goes up with acceleration \[\therefore \]
\[T-mg=ma\Rightarrow T=ma+mg=m(a+g)\] \[=\frac{1}{20}(1.2+9.8)=\frac{11}{20}=0.55N\] (b)
F.B.D in the case of elevator goes up with deceleration of \[1.2m{{s}^{-2}}.\] \[T-mg=-ma\] \[T=-ma+mg=m(g-a)=\frac{1}{20}(9.8-1.2)\] \[=\frac{8.6}{20}=0.43N\] (c)
F.B.D in the case of elevator goes up with uniform velocity. \[T=mg=\frac{1}{20}\times 9.8=0.49N\] (d)
F.B.D in the case of elevator goes down with acceleration of\[1.2m{{s}^{-2}}.\] \[mg-T=ma\Rightarrow mg-ma=T\] \[\Rightarrow T=m(g-a)=\frac{1}{20}(9.8-1.2)=0.42N\] (e)
F.B.D in the case of elevator goes down with deceleration of \[1.2m{{s}^{-2}}.\] \[mg-T=-ma\] \[\Rightarrow mg+ma=T\Rightarrow T=m(g+a)\] \[=\frac{1}{20}(9.8+1.2)=0.55N\] (f)
F.B.D in the case of elevator goes down with uniform velocity. \[mg-T=0\,\,\Rightarrow =mg=\frac{1}{20}\times 9.8=0.49N\]
You need to login to perform this action.
You will be redirected in
3 sec