(b) Explanation Given information: Two blocks are sliding down. Coefficient of kinetic friction between block of mass m 1 and surface is μ 1 . Coefficient of kinetic friction between block of mass m 2 and surface is μ 2 . Formula Used: According to Newton’s second law of motion: F n e t = m a Where, m is the mass and a is the acceleration. Calculation: Free body diagram is − Applying Newton’s 2 n d law at block 1 − N − m 1 g cos θ = 0 ... ( 1 ) T + m 1 g sin θ − f 1 = m 1 a T + m 1 g sin θ − μ 1 N = m 1 a ... ( 2 ) From (1) and (2) T + m 1 g cos θ − μ 1 m 1 g cos θ = m 1 a ... ( 3 ) Applying Newton’s 2 n d law at block 2 − N − m 2 g cos θ = 0 ... ( 4 ) m 2 g sin θ − T − f 2 = m 2 a m 2 g sin θ − T − μ 2 N = m 2 a ... ( 5 ) From equation (4) and (5); m 2 g sin θ − T − μ 2 m 2 g cos θ = m 2 a ..

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Chapter 5, Problem 52P

(b)

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Tension in the string joining two blocks when both blocks having equal coefficients of kinetic friction.

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Chapter 5, Problem 51P

Chapter 5, Problem 53P

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Tension in the string joining two blocks when both blocks having equal coefficients of kinetic friction.

Solution Summary: The author explains Newton's second law of motion and the formula used to calculate the acceleration of blocks.

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Tension in the string joining two blocks when both blocks having equal coefficients of kinetic friction.

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