Math  /  Algebra

Question33. (III) A 4.0kg4.0-\mathrm{kg} block is stacked on top of a 12.0kg12.0-\mathrm{kg} block, which is accelerating along a horizontal table at a=5.2 m/s2a=5.2 \mathrm{~m} / \mathrm{s}^{2} (Fig. 5-43). Let μk=μs=μ\mu_{\mathrm{k}}=\mu_{\mathrm{s}}=\mu. (a) What minimum coefficient of friction μ\mu between the twó blocks will prevent the 4.0kg4.0-\mathrm{kg} block from sliding offर्थ( (b)(b) If μ\mu is only half this minimum value, what is the acceleration of the 4.0kg4.0-\mathrm{kg} block with respect to the
FIGURE 5-43 Problem 33.

Studdy Solution
Calculate the acceleration of the 4.0 kg block if μ\mu is half the minimum value.
If μ\mu is half, then:
μ=0.5312=0.2655 \mu = \frac{0.531}{2} = 0.2655
The new frictional force f f' is:
f=μN=0.265539.2N=10.4N f' = \mu \cdot N = 0.2655 \cdot 39.2 \, \text{N} = 10.4 \, \text{N}
The net force on the 4.0 kg block is the difference between the force required to accelerate it and the frictional force:
Fnet=maf=20.8N10.4N=10.4N F_{\text{net}} = m \cdot a - f' = 20.8 \, \text{N} - 10.4 \, \text{N} = 10.4 \, \text{N}
The acceleration a a' of the 4.0 kg block is:
a=Fnetm=10.4N4.0kg=2.6m/s2 a' = \frac{F_{\text{net}}}{m} = \frac{10.4 \, \text{N}}{4.0 \, \text{kg}} = 2.6 \, \text{m/s}^2
The minimum coefficient of friction μ\mu is 0.5310.531, and if μ\mu is half this value, the acceleration of the 4.0 kg block with respect to the 12.0 kg block is 2.6m/s22.6 \, \text{m/s}^2.

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