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Math  /  Algebra

Question33. (III) A 4.0−kg4.0-\mathrm{kg} block is stacked on top of a 12.0−kg12.0-\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 4.0−kg4.0-\mathrm{kg} block from sliding offर्थ( (b)(b) If μ\mu is only half this minimum value, what is the acceleration of the 4.0−kg4.0-\mathrm{kg} block with respect to the
FIGURE 5-43 Problem 33.

Studdy Solution

STEP 1

1. The system consists of two blocks: a 4.0 kg block on top of a 12.0 kg block.
2. The 12.0 kg block is accelerating at a=5.2 m/s2 a = 5.2 \, \text{m/s}^2 .
3. The coefficient of kinetic friction (μk\mu_k) is equal to the coefficient of static friction (μs\mu_s) and is denoted by μ\mu.
4. We need to find the minimum coefficient of friction μ\mu to prevent the 4.0 kg block from sliding.
5. If μ\mu is half of this minimum value, we need to find the acceleration of the 4.0 kg block with respect to the 12.0 kg block.

STEP 2

1. Analyze the forces acting on the 4.0 kg block.
2. Determine the minimum coefficient of friction μ\mu.
3. Calculate the acceleration of the 4.0 kg block if μ\mu is half the minimum value.

STEP 3

Analyze the forces acting on the 4.0 kg block.
The force of friction f f is what prevents the 4.0 kg block from sliding off. This force is given by:
f=μ⋅N f = \mu \cdot N
where N N is the normal force, which is equal to the weight of the 4.0 kg block:
N=m⋅g=4.0 kg⋅9.8 m/s2=39.2 N N = m \cdot g = 4.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N}

STEP 4

Determine the minimum coefficient of friction μ\mu.
The frictional force must be equal to the force required to accelerate the 4.0 kg block at 5.2 m/s2 5.2 \, \text{m/s}^2 :
f=m⋅a=4.0 kg⋅5.2 m/s2=20.8 N f = m \cdot a = 4.0 \, \text{kg} \cdot 5.2 \, \text{m/s}^2 = 20.8 \, \text{N}
Set the frictional force equal to the required force:
μ⋅39.2 N=20.8 N \mu \cdot 39.2 \, \text{N} = 20.8 \, \text{N}
Solve for μ\mu:
μ=20.8 N39.2 N=0.531 \mu = \frac{20.8 \, \text{N}}{39.2 \, \text{N}} = 0.531

STEP 5

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′=μ⋅N=0.2655⋅39.2 N=10.4 N 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=m⋅a−f′=20.8 N−10.4 N=10.4 N 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=10.4 N4.0 kg=2.6 m/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 2.6 m/s22.6 \, \text{m/s}^2.

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