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Ch 09: Work and Kinetic Energy

Chapter 9, Problem 9

The energy used to pump liquids and gases through pipes is a significant fraction of the total energy consumption in the United States. Consider a small volume V of a liquid that has density p. Assume that the fluid is nonviscous so that friction with the pipe walls can be neglected. (a) An upward-pushing force from a pump lifts this volume of fluid a height h at constant speed. How much work does the pump do?

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Hi, everyone. In this practice problem, we are being asked to calculate the work done by an electric water pump used to lift water into a tank. We will have salty water with a density of row equals to 1010 kg per meter cube. And the pump will raise a volume element of fee equals to 250 liter into the tank. The water is going to be a non fiscus liquid that will not experience any sort of friction from the walls of the pipe. And the pump exerts an upward force on the water, lifting the water to the tank for a total height of h equals to m. Using all of this information, we're being asked to determine the total work done by the pump and the options given are a, 100 and 98 mega Joules B 100 and kilojoules and C 3.17 kilojoules and last but not least d uh 1.98 mega Joules. So in this problem, we will ignore the changes in volume and the shape of the water. And we will also ignore the viscosity for water to actually move up the pipe at a constant speed. We know that the work done by the electric water pump must be equal to the work done by gravity but in the opposite direction. So in this case, we know that the total, the only two work acting upon our system are the pump and the gravity, we denote the pump with a subscript of P and gravity with a subscript of G. So in this case, our total work done or our net work done W net is going to equals to W G plus W P. From the work energy theorem, we know that W net will equals to the change in kinetic energy. And from the fact that the work done by the gravity is equal to and in opposite direction to the work done by the water pump, we know that our W net will actually equals to zero. So then therefore four R W G will then equals to negative W P or vice versa. W P will equals to negative W G. We want to recall that R W G or the work done by gravity can be calculated equals to negative M G H where the mass mass can be substituted with row multiplied by P or volume. So we want to substitute the mass into our W G equation so that our W G will then equals to negative row P G H. And that is done or obtained by substituting our M into the W G equation just like. So next, we want to actually substitute our W G equation that we found into R W P so that we can eventually calculate the total work done by our pump. So W B equals negative W G and substituting R W G that means W P will equals negative of negative row V G H, the two negative will cancel out. So R W P will then just be row multiplied by V multiplied by G multiplied by H. So we actually can start substituting all of our information into this equation right here. So the W B will then be row is going to be 1000 and 10 kg per meter cube. And then our volume is given in the problem statement to be 250 liter. And we wanna convert that into meter cube into si that will be multiplied that by one m cube divided by 1000 liter just like so and then we wanna multiply that by our gravitational acceleration which is 9. m per second squared. And we wanna multiply all of that with our H or the height which is a meter, all of that uh will then sum up or multiply it up to 10, 100 and 97, kg multiplied by meter squared divided by a second squared or that equals to 100 and 97, Joles which will rounding it up, which will be equals to 100 and 98 kg Joles. So 100 and 98 kilojoule is actually going to be the answer to this practice problem and that will essentially be corresponding to option B. So option B is going to be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topic and that'll be it for this one. Thank you.