Efficiency of an inclined plane

[pfk123]

Homework Statement

List 2 major factors result in an efficiency of less than 100% when using an inclined plane to raise a mass

Note - this is a grade 11 lab so don't have answers that are too advanced please :D

Homework Equations

eff = eout/ein x 100%
... = mghf/fd x 100%

The Attempt at a Solution

I listed friction as one with the explanation: friction (thermal energy) when the block with mass was being pulled up the inclined plane, it experiences a force of kinetic friction opposite to the applied force. This is a result of the chemical bonding between surfaces. Since W = Fd, an increase in applied force was needed at first to overcome static friction, then kinetic friction after. An increase in applied force will result in an increase of work as they are proportional and the displacement did not change. Since friction can never be fully eliminated and is usually not the useful energy, there will be efficiency that is less than 100%.

I have no clue what the 2nd major factor is though. It was not sound as it was a small block and didn't make noise going up the ramp. Should I just use sound as my second factor?

 

[PhanthomJay]

pfk123, welcome to PF!
Your explanation of friction as a major factor in efficiency loss is very good.

Regarding the 2nd factor, you mentioned sound. There is also heat generated. And other forms of energy. But all these forms of energy are a result of the work done by friction, so these losses are sort of tied together as the same factor. There is another factor that relates to the magnitude of the friction force that must be overcome. It has to do with the ramp geometry. Can you think what it might be? And there is another large factor relating to the friction force magnitude...think about the ramp surface... both these factors are subfactors of your friction explanation