Math Problem Statement

A 11kg block is placed on top of a 30 kg block. A force of 367 N is applied to the right of the lower block, and the upper block slips on the lower block. The coefficient of kinetic friction between the upper block and the lower block is 0.2 and the coefficient of kinetic friction between the lower block and the floor is 0.58. How big would the coefficient of static friction between the upper and lower block have to be so that the upper block would not slip on the lower block?

Solution

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law of Motion
Friction Forces
Statics and Dynamics

Formulas

F = ma (Newton's second law)
f_s = μ_s N (static friction force)
f_k = μ_k N (kinetic friction force)
a = F_net / m_total (acceleration)

Theorems

Newton's Second Law
Frictional force relations

Suitable Grade Level

Grades 11-12 (Physics)