A block with mass m 1 = 9.2 kg is on an incline with an angle = 39° with respect

Question

A block with mass m1 = 9.2 kg is on an incline with an angle = 39° 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 = 0.35 and s = 0.385.

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

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

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 = 0.16 m from its unstretched length.

4)

Now a new block with mass m2 = 16.4 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 39° 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 = 0.385 x 9.2 x 9.81 x Cos 39 = 27 N

Since µk = 0.35, then sliding frictional force is:
27 x 0.35 / 0.385 = 24.5 N

Resultant Force down plane = mg Sin – 24.5 = (9.2 x 9.81 x Sin 39) – 24.5 = 32.3 N

F = m x a
32.3 = 9.2*a
Acceleration, a = 32.3/9.2 = 3.51 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 = (9.2*9.81)*(sin39 – cos39*0.385)/0.16 = 186.2 N/m

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
=>(9.2+16.4) x tan39* = 9.2 x 0.385 + 2s x 16.4
=> 2s = 1.048

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