What Modifications Can Boost Ramp Efficiency Without Height Adjustment?

AI Thread Summary
To boost ramp efficiency without height adjustment, reducing friction and air resistance is crucial. Enhancing the ramp's surface smoothness can minimize friction, while making the object more aerodynamic can decrease air resistance. The discussion emphasizes that efficiency cannot exceed 100%, as some energy is always lost. Participants suggest that achieving better efficiency involves optimizing these factors rather than altering the ramp's height. Overall, improving surface and object design can lead to increased efficiency in the ramp setup.
iRamie
Messages
19
Reaction score
1

Homework Statement


We have a lab in which there is a ramp with one end on the floor and the other end is on top of a few textbooks. Therefore, it has a slope, and an object is lifted using the ramp.
What can we do to this set-up to increase the efficiency, OTHER THAN INCREASING THE HEIGHT.

Homework Equations



Efficiency = Eout / Ein
Eout = mgh
Ein = Work = FD

The Attempt at a Solution

 
Physics news on Phys.org
If the kinetic energy is unchanged during the process, the Eout equals the Ein. So the efficiency is 1 according to your definition for it. Can you get better efficiency than 1?
 
It has a slope? The efficiency can't be 100%? Some work has to be lost right?
 
Well if there is friction and air resistance, perhaps you could do something to decrease these factors... perhaps make the surface more smooth and the object more aerodynamic. Try to think for yourself what might help to increase efficiency. I think you could have come to these same conclusions.
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top