A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp,

Question

A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp, inclined at 25 degress with the horizontal.During this slide 20 J of energy is lost due to friction. What is the kinetic friction force on the crate? What is the speed of the crate at the bottom of theramp? A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp, inclined at 25 degress with the horizontal.During this slide 20 J of energy is lost due to friction. What is the kinetic friction force on the crate? What is the speed of the crate at the bottom of theramp?

Explanation / Answer

A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp, inclined at 25 degress with the horizontal.During this slide 20 J of energy is lost due to friction. What is the kinetic friction force on the crate?

work done due to opposing kinetic friction=20J=20**cos25*5
so kinetic friction=0.2206
What is the speed of the crate at the bottom of the ramp?

acceleration down theplane=g(sin25-0.2206cos25)=2.18246m/sec2


so speed at the end of ramp=0+2*2.1824*5=4.6716806398m/sec

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