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InterviewSolution
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A train is moving with a speed of `90 kmh^(-1)` passenger X inside the train displaces his 40 kg, luggage slowly an the floor through 1 m in 10 s. Coefficient of friction of the floor of the train is 0.2. Find the work done by this passenger X and the luggage as seen by (i) a follow passenger Y (ii) a person on the ground [Take ,g=10 `ms^(-1)`]. |
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Answer» Given, speed of the train, `v=90kmh^(-1) =(90xx1000)/(60xx60)=25 ms^(-1)` (i) Displacement of the luggage with respect to the train,S=1 m. As luggage is displaced slowlt, the force applied on it must be same as frictional force on it by the floor `f=mu" mg"=(0.2)xx40xx10=80 N` Work done by the passenger X as seen by fellow passenger Y is `W =F_(s)=(80)(1)=80 J` (ii) The luggage is dispalced for 10 s. Therefore distance moved by train with respect to ground during this interval is `s_(0)=25xx10=250 m` Therefore work done by passenger X on the luggage as seen by a person on the ground is `W_(G)=F(s+s_(0))=(80)(1+250)=(251)(80)=20.08 kJ` |
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