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)`].

Explanation

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`