Question

Block on moving inclined plane calcute accleleration

Answer 1

The force due to gravity, Fg, will be resolved into two forces:

Fn=Fg∗Cos(θ) (normal force)

Ft=Fg∗Sin(θ) (tangential force)

The tangential force points down the plane, and causes the object to accelerate. The normal force points into the plane, and causes friction, which opposes the tangential force.

The frictional force, if the object is not moving on the plane, will be: Ff(static)=Fn∗μs where μs is the static coefficient of friction, which depends on the surfaces of the object and the ramp. If Ff(static) is greater than or equal to Ft, then the object won’t move.

If Ff(static)<Ft, then the object will accelerate. Once the object’s velocity rises above 0 (basically instantly), the frictional force becomes: Ff(kinetic)=Fn∗μk, where μk is the kinetic coefficient of friction, which is typically less than μs, so it’s unlikely that Ff will be greater than Ft in this case. The total force on the block, pointing down the ramp, will be Ft−Ff(kinetic). So the acceleration will be a=Ft−Ffm.

If we use the equations above, we can substitute Fg∗Sin(θ) for Ft, and Fn∗μk for Ff. That gives us:

a=Fg∗Sin(θ)−Fn∗μkm

We can now substitute Fg∗Cos(θ) for Fn, giving us:

a=Fg∗Sin(θ)−Fg∗Cos(θ)∗μkm

a=Fg∗Sin(θ)−μkCos(θ)m

Since we know that acceleration due to gravity is 9.8 m/s^2,

I hope this will help you

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