A block with mass m 1 = 9.2 kg is on an incline with an angle = 33° with respect

Question

A block with mass m1 = 9.2 kg is on an incline with an angle = 33° with respect to the horizontal. For the first question there is no friction between the incline and the block.

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

2)Now with friction, the acceleration is measured to be only a = 4.06 m/s2. What is the coefficient of kinetic friction between the incline and the block?

3)

To keep the mass from accelerating, a spring is attached with spring constant k = 164 N/m. What is the coefficient of static friction if the spring must extend at least x = 16 cm from its unstretched length to keep the block from moving down the plane?

4)

The spring is replaced with a massless rope that pulls horizontally to prevent the block from moving. What is the tension in the rope?

5)

Now a new block is attached to the first block. The new block is made of a different material and has a coefficient of static friction = 0.97. What minimum mass is needed to keep the system from accelerating?

Explanation / Answer

1) a = g*sin
on a frictionless incline. Here, a = 9.8*sin33 = 5.34 m/s²

2) When there is friction, a = g*(sin - µ*cos).
4.06 m/s² = 9.8m/s² * (sin33 - µ*cos33) ========> µ = 0.155

3) Now we can add another term:
a = g*(sin - µ*cos) - kx/m
Dropping units for ease,
0 = 9.8*(sin33 - µ*cos33) - 164*0.16/9.2 ======> µ = 0.302

4) Assuming µ = 0.302,
we still have the downslope acceleration g*sin
but now upslope we have the rope tension component (T/m)cos
and a friction component that is greater than the weight component alone because of the tension component perpendicular to the incline:
µ(gcos + (T/m)sin)). So
0 = g*sin - (T/m)*cos - µ*(g*cos + (T/m)*sin)
Plugging in numbers,
0 = 9.8*sin33 - (T/9.2)*cos33 - 0.302*(9.8*cos33 + (T/9.2)*sin33) ======> T = 26.2 N

5) I'll finally have to work with forces instead of accelerations.
For the first block:
a = g*(sin - µ*cos) - (T/m) assumes the string is now parallel to the incline
Since a = 0,
T = 9.2kg * 9.8m/s² * (sin33 - 0.302*cos33) = 26.3 N

Now for the second mass, then
a = g*(sin - µ*cos) + (T/M)
0 = 9.8*(sin33 - 0.97*cos33) + 26.3/M ======> M = 9.98 kg.

Hope this helps!

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