Lifting books - Potential Energy

Thread starter Ipos Manger
Start date Jun 20, 2011
Tags

Books Energy Lifting Potential Potential energy

AI Thread Summary

To calculate the minimum work required to stack six books, each weighing 30N and 4cm thick, the total height of the stack must be considered. The work done against gravity is determined by the formula W = F × d, where F is the force (weight of the books) and d is the distance moved. The distance for each book varies, with the bottom book requiring no lift and the top book needing to be lifted the full height of the stack. The total work can be calculated by summing the individual work done for each book as they are lifted to their respective heights. Understanding these principles is essential for solving the problem effectively.

Jun 20, 2011

Ipos Manger

I need help on this proble:

A student has six books, all of them with a 4cm thickness and a weight of 30N. ¿What would be the minimum work to put the books in one pile, if the six books are in a table?

Physics news on Phys.org

Jun 20, 2011

Pengwuino

Gold Member

Follow the template you deleted. You need to show some effort before people will help you.

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »

Thread 'Minimum mass of a block'

Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...

View full post »