A crate of mass 15kg is placed on a plane inclined at 60 degrees to the horizontal. If the crate slides down with a constant speed, calculate the coefficient of kinetic and the magnitude of the frictional force acting on the crate (g=10m/s)
Answers
Answer:
1.732 (4 s.f)
Explanation:
We will do forces acting on the horizontal plane since it is sliding down.
Newton's Second Law suggests ∑Fₓ=maₓ, therefore we can resolve the force acting in the horizontal direction down the plane as (mg sin 60) and force acting downwards towards the centre of the Earth as (mg cos 60), mg being weight downwards. We also have friction, which is μN, where N is our force acting downwards (mg cos 60), and μ being the coefficient of kinetic friction which acting backwards the crate, so that force has to be subtracted from the force acting forwards.
Subbing our values into ∑Fₓ=maₓ :
mgsin 60 - μ(mg cos 60)= 0 <-- (equals to zero because constant velocity means no acceleration)
our mass,m is 15 and our gravitational force is 10m/s
Therefore if we rearrange, (15)(10)sin60 = μ (15)(10) cos 60
And finally, μ= sin 60/cos 60 = 1.732
Answer:
Explanation:
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