A block with mass m 1 = 9 kg is on an incline with an angle = 30° with respect t

Question

A block with mass m1 = 9 kg is on an incline with an angle = 30° with respect to the horizontal. For the first question there is no friction, but for the rest of this problem the coefficients of friction are: k = 0.27 and s = 0.297.

1)When there is no friction, what is the magnitude of the acceleration of the block?

__________________m/s2

2)Now with friction, what is the magnitude of the acceleration of the block after it begins to slide down the plane?

_________________m/s2

3)To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.12 m from its unstretched length.

____________________N/m

4)

4)Now a new block with mass m2 = 16.7 kg is attached to the first block. The new block is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating?

Explanation / Answer

Assume that 30° is the steepest angle at which block will not slide down on its own.
This is then the limiting friction
F = µ R = µ m g Cos = 0.297 x 9x 9.81 x Cos 30 = 22.71 N

Since µk = 0.27, then sliding frictional force is:
22.71 x 0.27 / 0.297 = 20.645N

Resultant Force down plane = mg Sin - 20.645 = (9 x 9.81 x Sin 30) - 20.645

Force = 23.455N

F = m x a
23.455 = 9 a
Acceleration, a = 2.606 m/s^2

3.Since the block is not moving, we use s rather than k. That means the force of friction (Ff) is given by this inequality:

Ff (Fn)s
(Where Fn is the normal force.) For a block resting on a slope, the normal force is (mg)cos, so:

Ff (mg)coss

The components of force parallel to the slope are:
1. The upslope pull of the spring (= kx, where k is the spring constant)
2. The upslope force of friction (= Ff)
3. The downslope component of the block's weight (= (mg)sin)

These add up to zero (because there's no acceleration). That means (choosing "upslope" as the positive direction):

kx + Ff - (mg)sin = 0

Solve for Ff:
Ff = (mg)sin - kx

Combine with the inequality above:
(mg)sin - kx (mg)coss

Solve for k:
k (mg)(sin - coss) / x

4.By F(net) = F(gravity) - F(friction)
=>if there is no acceleration=>F(net) = 0
=>F(gravity) = F(friction)
=>m1sin + m2sin = 1s x m1cos + 2s x m2cos
=>(m1+m2) tan = m11s + m22s

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