(a) Explanation Given information: The weight of the block A ( W A ) is 0.75 lb . The weight of the block B ( W B ) is 0.5 lb . The distance between the block B and block C ( x ) is 0.15 in . or 0.0125 ft . Assume the acceleration due to gravity ( g ) is 32.2 ft / s 2 . Calculation: Calculate the mass of block A ( m A ) using the relation: m A = W A g Substitute 0.75 lb for W A and 32.2 ft / s 2 for g . m A = 0.75 32.2 = 0.0233 lb ⋅ ft / s 2 Calculate the mass of block B ( m B ) using the relation: m B = W B g Substitute 0.5 lb for W B and 9.81 m / s 2 for g . m B = 0.5 32.2 = 0.015528 lb ⋅ ft / s 2 Calculate the force ( F ) using the formula: F = k x 2 Substitute 0.0125 ft for x . F = k 0.0125 2 = 6400 k Calculate the constant ( k ) by considering the vertical equilibrium condition: ∑ F = 0 − W A − W B + F = 0 Substitute 0.75 lb for W A , 0.5 lb for W B and 6400 k for F . − 0.75 − 0.5 + 6 , 400 k = 0 k = 195.3 × 10 − 6 lb ⋅ ft 2 Here, the initial kinetic energy ( T 1 ) is zero due to zero initial velocity. Calculate the final kinetic energy ( T 2 ) using the formula: T 2 = 1 2 m B v max 2 Substitute 0.015528 lb ⋅ ft / s 2 for m B . T 2 = 1 2 ( 0.015528 lb ⋅ ft / s 2 ) v max 2 = 7.7639 × 10 − 3 v max 2 lb ⋅ ft Calculate the work done ( U 1 − 2 ) G by the magnetic force and weight using the formula: ( U 1 − 2 ) = ∫ 0.0125 x ( F − W B ) d x Substitute k x 2 for F and 0 (b) To determine Find the maximum acceleration ( a B ) of B .

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ISBN: 9781260663778

Author: BEER

Publisher: MCG

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Chapter 13.1, Problem 13.32P

(a)

To determine

Find the maximum velocity (vmax) of B.

(b)

To determine

Find the maximum acceleration (aB) of B.

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Chapter 13.1, Problem 13.31P

Chapter 13.1, Problem 13.33P

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Find the maximum velocity ( v max ) of B .

Solution Summary: The author calculates the mass of the block A (w_A) using the relation: