I can not solve this problem. please help. My friends told it's easy.?

Question

A block is resting on a ramp. You can change the inclination angle (theta) by raising one end of the ramp. The block has a mass m = 9.5 kg. At the interface between the ramp and the block, the coefficient of static friction is ms = 0.45, and the coefficient of kinetic friction is mk = 0.315.
If you raise the end of the ramp very slowly so that you can assume a = 0 for the block, at what angle will it slip?

Also,If it slips at that angle, what will the magnitude of its acceleration be?

And, if you can, How long does it take from the time it slips to travel a distance d = 0.90 m down the ramp?

Thank you so much.

Answer 1

Answer to your problem.
tan θ = μ = 0.45
θ = 24.22°
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Since it starts from rest
s = 0.5 a t^2
0.9 = g [sin θ - μk cos θ] t^2.
0.9 = 9.8 [sin 24.22- 0.315 cos 24.22] t^2
t = 0.864 s.
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DETAIL
Let θ be the inclination, at which the block just begins to slide.

mg sin θ is the force that is trying to pull the block down the plane.
The static friction opposes this.
Static friction is μ times the normal reaction which is mg cos θ.
The block is at rest
mg sin θ = μ mg cos θ
tan θ = μ
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The block is at rest when tan θ = μ
μ is the coefficient of static friction.
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When the angle is increased from θ by a very small angle it will begin to slide down.
Once it has begun to slide the friction will be reduced to μ[kinetic] The plane is again brought to angle θ
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mg sin θ >μk mg cos θ
The force acting on the block is now
mg [sin θ - μk cos θ]

The acceleration is now = g [sin θ - μk cos θ]
===========================

Answer 2

Resolve the downward force of gravity into components, one parallel to and one perpendicular to the ramp. The block will start to slip when the ratio of these becomes 0.45. Once the block is moving, the parallel component, reduced by the perpendicular component times 0.315, provides an acceleration. The usual formula can then be used to see what happens thereafter. Note that at no point in this mess does the particular mass of the block affect anything.

Answer 3

At sinθ = 0.45 the component of the weight parallel to the ramp will exactly balance the static friction force. At 26.74369° the box starts to slide and sliding friction takes over, yielding an acceleration of 0.135 m/s^2
(1/2)(0.315)t^2 = 0.90 m
t^2 ≈ 13.333 s^2
t ≈ 3.65 s