Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
6
Work and Energy
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6-1: Work, Constant Force
6-3: Kinetic Energy; Work-Energy Principle
6-4 and 6-5: Potential Energy
6-6 and 6-7: Conservatino of Mechanical Energy
6-8 and 6-9: Law of Conservation of Energy
6-10: Power

Problem 9
A
a) Fup=1.10gMF_{up}=1.10gM
b) W=1.1gMhW=1.1gMh
Giancoli 6th Edition, Chapter 6, Problem 9 solution video poster
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VIDEO TRANSCRIPT

Lets draw the helicopter first and then we'll be able to put some force vectors on there and we'll know how to calculate the net force relate the to acceleration. So there is the helicopter and we have the force up provided by the spinning rotors and gravity is pulling it down, gravity is going to be a little shorter because it has a net force up; its accelerating upwards. Let's write down the net force, it's the up force minus the down force, which is gravity, and that net force is equal mass times acceleration and we are told that's 'M', which is the mass of this helicopter times 0.1 'g's' so that's the acceleration. Then we can solve for the up force so 'Fup' equals 0.1 'g' times mass of the helicopter plus 'Mg' that's gravity. Then we can simplify that a little bit by factoring up the capital 'M' 0.1 'g' plus 'g' and we'll make that one look a little nicer even 1.10 'g' times 'M' that's what the force up is

For part B we'll find the work done by that force and it's going to be that force times the distance over which it acts which is 1.10 'g' times 'M', that's the up force and the distance that it [enacts] is the height 'h' that the helicopter is raised. So this is the work done by the rotors; the force created by the rotors.

COMMENTS
By smoothtforu on Tue, 10/26/2010 - 2:35 PM

the site is not working. It keeps saying server not available

By Mr. Dychko on Tue, 10/26/2010 - 6:30 PM

Yes, sorry about that Smoothforu, we resized our server this morning to handle more visitors. The site was unresponsive for about 5min, but now it's a little bit faster.

By trandzuyhieu on Sat, 7/9/2016 - 4:01 AM

Can you explain how M(0.10g) is equal to Force up - Force mg = ma?
I get the equation: F=ma and sum of F =Fup-Fmg=ma. I don't get how M(0.10g) is equal to all that.

Thanks

By Mr. Dychko on Sat, 7/9/2016 - 9:32 PM

Hi trandzuyhieu, thanks for the question. M(0.10g)M(0.10g) replaces the mama in Fupmg=maF_{up} - mg = ma. Data from the question say the helicopter mass is MM, and it's acceleration is (0.10g)(0.10g), so we substitute for each of these in mama: MM replaces mm, and (0.10g)(0.10g) replaces aa.

All the best,
Mr. Dychko

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