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A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration\[\text{1}.\text{2 m}/{{\text{s}}^{\text{2}}}\], (b) goes up with deceleration 1.2 m/s 2, (c) goes up with uniform velocity, (d) goes down with acceleration\[\text{1}.\text{2 m}/\text{s}{{\text{ }}^{\text{2}}}\], (e) goes down with deceleration \[\text{1}.\text{2 m}/{{\text{s}}^{\text{2}}}\] and (f) goes down with uniform velocity.         

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. \[mg-T=0\,\,\Rightarrow =mg=\frac{1}{20}\times 9.8=0.49N\]