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two identical 2.0 kg blocks are stacked as shown in figure p5.71. the bottom block is free to slide on a frictionless surface. the coefficient of static friction between the blocks is 0.35. what is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

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Final answer:

The maximum horizontal force that can prevent the upper block from slipping on the lower block is calculated using the coefficient of static friction and the weight of the upper block. It is determined to be 6.86 N.

Step-by-step explanation:

To determine the maximum horizontal force that can be applied to the lower block without the upper block slipping, we need to calculate the maximum static friction force between the two blocks. The maximum static friction force (F_max) is given by the equation F_max = μ_s × N, where μ_s is the coefficient of static friction and N is the normal force. Since the blocks are stacked, the normal force is equal to the weight of the upper block, which is the mass of the block multiplied by the acceleration due to gravity (g).

The calculation involves the following steps:

  1. Calculate the weight of the upper block: Weight = mass × g.
  2. Calculate the maximum static friction force: F_max = μ_s × Weight.

The mass of the upper block is 2.0 kg, the acceleration due to gravity (g) is approximately 9.8 m/s², and the coefficient of static friction (μ_s) is 0.35. Thus:

  1. Weight = 2.0 kg × 9.8 m/s² = 19.6 N.
  2. F_max = 0.35 × 19.6 N = 6.86 N.

The maximum horizontal force that can be applied to the lower block without the upper block slipping is 6.86 N.

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