Ram and Ali have been fast friends since childhood. Ali neglected studies and now has no means to earn money other than a channel whereas Ram has become an engineer. Now both are working in the same factory. Ali uses camel to transport the load within the factory.
Due to low salary and degradation in health of camel, Ali becomes worried and meets his friend Ram and discusses his problems. Ram collected some data and with some assumptions concluded the following:
i. The load used in each trip is `1000kg` and has friction coefficient `mu_k=0.1` and `mu_s=0.2`.
ii. Mass of camel is `500kg`.
iii. Load is accelerated for first `50m` with constant acceleration, then it is pulled at a constant speed of `5ms^-1` for `2km` and at last stopped with constant retardation in `50m`.
iv. From biological data, the rate of consumption of energy of camel can be expressed as `P=18xx10^3v+10^4Js^-1` where P is the power and v is the velocity of the camel. After calculations on different issues, Ram suggested proper food, speed of camel, etc. to his friend. For the welfare of Ali, Ram wrote a letter to the management to increase his salary.
(Assuming that the camel exerts a horizontal force on the load):
Maximum power transmitted by the camel to load is

Options

A. (a) `2.1xx10^6J`
B. (b) `4.22xx10^7J`
C. (c) `2.22xx10^4J`
D. (d) `4.22xx10^6J`

Answer

B

Explanation

The time of travel in accelerated motion = time of travel in retarded motion.
`t_(AB)=t_(CD)=v/a=(5)/(0.25)=20s`
Now time for uniform motion `=T_(BC)=2000//5=400s`
Total energy consumed is
`underset0overset440int Pdt= underset0overset20 int [18xx10^3v+10^4]dt+underset20overset420int[18xx10^3xx5+10^4]dt+underset420overset440int[18xx10^3v+10^4]dt`
Putting `vdt=dx` and changing limits approximately, we get `vdt=dx`
It becomes
`underset0overset50int18xx10^3dx+[10^4t]_0^20+10^5[420-20]+underset(2050)overset(2100)int 18xx10^3dx+[10^4]_(420)^(440)`
`=18xx10^3xx50+10^4[20]+10^5xx400+18xx10^3[50]+10^4[20]J`
`=4.22xx10^7J`