An airport worker pushes a 150.0 kg crate up a ramp that is 3.50 m high and 8.00 m long into the cargo bay of an airplane. He exerts a 700.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Include all work and a free-body diagram!

***Please explain? Thank you:)

Question

An airport worker pushes a 150.0 kg crate up a ramp that is 3.50 m high and 8.00 m long into the cargo bay of an airplane. He exerts a 700.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Include all work and a free-body diagram!

***Please explain? Thank you:)

anonymous · Answer

draw your diagram first

anonymous · Answer

I am not sure what the diagram would or should look like :(

phi · Answer

draw the ramp, and put in the numbers, and the crate

anonymous · Answer

okay, sort of like this? |dw:1426875836143:dw|

phi · Answer

here are examples http://www.interactagram.com/physics/dynamics/MechanicalAdvantage/ramp/

anonymous · Answer

ohh so it should be this? |dw:1426875921984:dw|

anonymous · Answer

|dw:1426875974823:dw|

anonymous · Answer

so mechanical advantage would be 8.00/3.50 = 2.285714286 and force required is 1/2*150=75? or would that be the 700 N that they give in the problem?

anonymous · Answer

oh oops, do i need to find the force like that? Or is that not necessary here?

phi · Answer

do you have a formula in your notes for efficiency?

anonymous · Answer

i think so... is it this?

efficiency=work output/work input * 100% ?

phi · Answer

and what is work input for this problem ?

anonymous · Answer

work input is 700.0 N ?

phi · Answer

700 N is a force

anonymous · Answer

ohh okay :/ erm not sure what the work input is? would that be the 150kg?

phi · Answer

I would think 700 * 8 ?

anonymous · Answer

ohh okay so the work input = 5600? 

so efficiency =work output/5600 * 100% ?

phi · Answer

now we need work out

anonymous · Answer

okay, how can we find that? :O

phi · Answer

probably weight * height lifted ?

anonymous · Answer

okay, so 150*3.5 = 525? 

so efficiency = 525/5600 * 100% = 0.09375 * 100% = 0.09375? :/

phi · Answer

we need to get the units correct
i.e.  N-m
N = kg-m/sec^2
so  we need to get  kg-m^2/sec^2

it sounds like we want the force due to gravity:  F= mass * acceleration=  m g

anonymous · Answer

okay, so efficiency = 0.09375 kg-m^2/sec^2 ? :/ 


and so F=150*a?

phi · Answer

in other words,   work = mass kg * 9.8 m/sec * 3.5 meters

anonymous · Answer

ohhh
okay so F=150kg * (9.8 m/sec) * (3.5 m) ?

phi · Answer

yes, that looks better

phi · Answer

but it's not F , its work = F * distance

anonymous · Answer

okay! so F=5145? :/

phi · Answer

work

anonymous · Answer

oh, oops work=5145 kg-m/sec^2

anonymous · Answer

?

phi · Answer

yes, that is work out

anonymous · Answer

ahh okie! so now we have this?

efficiency = 5145/5600 * 100% = 0.91875 or 91.875% efficiency?

phi · Answer

yes

anonymous · Answer

ahh okkie!! :) thanks so much!! :D