EK Physics question - why are they using Work to solve this problem?

[delempicka]

Some guy wants a ramp to reduce the force necessary to lift boxes to the top of a .5 m step. If there is only enough room for a 4m ramp, what is the maximum factor by which the lifting force could be reduced?

Ans. Key states: The work done will remain the same. W=Fd. the distance is increased from .5 to 4; this is an increase by a factor of 8. since work remains constant, force must decrease by a factor of 8.


My ans/reasoning -

I am assuming when they say a 4m ramp they mean the hypoteneuse is 4m...



the point of an incline plane is to reduce the amt. of Force necessary to move an object from A (low) to B (high). w/o incline - F=mg. incline works by decreasing the amt. of accel due to gravity a body experiences in transit, via the addition of a horizontal distance component. thus, less blah blah blah

...so f=mgsinq,
sinq=o/a=(.5/4)
=1/8

force goes down by 8.

they bring work into matter and say work is constant...aren't they talking about a step which would be vertical - and by f=mg cosq - equal 0.

is this reasoning wrong? what's up? am i getting the right ans. via the wrong route?
thx

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[thatscorrect7]

it has to do with mechanical advantage

effort distance/resistance distance

Work is the same because it's conservative.

A machine only changes the force and not the work.

If you lift a book to 4 meters and you have a ramp that is 4 meters high then the potential energy at the top is the same.

 

[delempicka]

thx for ans...

but when they say "there is only room for a 4m ramp" are they saying that the incline plane is 4 m (hypoteneuse) or it is 4m from the step to the point where the plane meets the incline...

 

[TheBoondocks]

thx for ans...

but when they say "there is only room for a 4m ramp" are they saying that the incline plane is 4 m (hypoteneuse) or it is 4m from the step to the point where the plane meets the incline...

Hypotenuse is 4 m and height is .5 like you said. MA