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# Ram and Ali have been fast friends since childhood. Ali neglected studies and now has no means to earn money other than a camel whereas Ram becomes an engineer. Now both are working in the same factory. Ali uses camels to transport the load within the factory. Due to low salary and degradation in the health of the camel, Ali becomes worried and meets his friend Ram and discusses his problems. Ram collected some data and some assumptions concluded the followings: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 $5m{s^{ - 1}}$ for $2km$ and at last stopped with constant retardation in $50m$.iv) From Biological data, the rate of consumption of energy can be expressed as $P = 18 \times {10^3}v + {10^4}J{s^{ - 1}}$ 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 friends. 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)The ratio of magnitude of work done by Camel on the load during accelerated motion to retarded motion is:

Last updated date: 13th Jun 2024
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Hint: First calculate the work done by the camel on the load during the accelerated motion by using the kinetic energy and similarly the work done by the camel on the load during the retarded motion.

Complete step by step solution:
In this question, we have given the load is $1000kg$ and friction coefficient ${\mu _k} = 0.1$ and ${\mu _s} = 0.2$.
Mass of the camel is $500kg$.
Hence, Work done by camel on load during the accelerated motion is, $\left( {\dfrac{1}{2}m{v^2} - 0} \right) - \left( { - {u_k}mgr} \right)$
[$r$ is retardation]
Now we will substitute the values in the above equation as,
$\therefore \dfrac{1}{2} \times 1000 \times {5^2} + \left( {0.1 \times 1000 \times 10 \times 50} \right) = 62500$
Now work done by the came on load during the retarded motion is, $\left( {0 - \dfrac{1}{2}m{v^2}} \right) - \left( { - {u_k}mgr} \right)$$= \left( {0.1 \times 1000 \times 10 \times 50} \right) - \dfrac{1}{2} \times 1000 \times {5^2} = 37500$

Hence the ratio will be $\Rightarrow \dfrac{{62500}}{{37500}} = 5:3$.