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A block with mass [tex]m_1 = 7.04 \, \text{kg}[/tex] is placed on top of a block with mass [tex]m_2 = 11.2 \, \text{kg}[/tex]. A horizontal force of [tex]99.1 \, \text{N}[/tex] is applied to the [tex]11.2 \, \text{kg}[/tex] block, causing it to slide. The [tex]7.04 \, \text{kg}[/tex] block is tied to the wall by a string. The coefficient of kinetic friction between all surfaces is [tex]0.246[/tex]. What is the acceleration of the system?

A. [tex]3.13 \, \text{m/s}^2[/tex]
B. [tex]5.39 \, \text{m/s}^2[/tex]
C. [tex]7.65 \, \text{m/s}^2[/tex]
D. [tex]10.12 \, \text{m/s}^2[/tex]

Answer :

Final answer:

The acceleration of the system of blocks is determined to be 2.32 m/s². This is computed using the coefficient of kinetic friction and Newton's second law, but it's not one of the provided choices, so the question or options might have an error.

Explanation:

The first step is to recognize that the friction between the blocks acts in opposite directions for each block. The friction force (F_f) can be computed by using the formula: F_f=μN, where μ is the coefficient of friction and N is the normal force. In this case, N=m1g for the top block and (m1+m2)g for the bottom block, where g is the acceleration due to gravity (9.81 m/s²). Hence, F_f1=μm1g and F_f2=μ(m1+m2)g. The net external force acting on the system is the applied external force (F) minus the frictional forces. Using Newton's second law (F=ma), the acceleration of the system is given by a=(F-F_f1-F_f2)/(m1+m2).

Plugging in the given values, we get: F_f1=0.246*7.04*9.81≈17.13 N, F_f2=0.246*(7.04+11.2)*9.81≈43.48 N, and the netexternal force=99.1-17.13-43.48=38.49 N. Thus, the acceleration of the system is a=38.49/(7.04+11.2)=2.32 m/s², which is not one of the listed options. It appears that the provided option choices are incorrect or there may be an error in the problem setup.

Learn more about Kinetic Friction here:

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