Explanation Given information: The weight of the block A and B are w A = 5 lb and w B = 15 lb . The spring constant ( k ) of the spring is 72 lb/in . Neglect the effect of friction. The compression ( e ) in the spring is 3 in . Calculation: Consider the mass of the block A and B are denoted by m A and m B . Consider the value of acceleration due to gravity as g = 32.2 ft/s 2 . Show the unit conversion of the mass of the block A and B as follows: m A = w A g = 5 lb 32.2 ft/s 2 = 0.15528 lb ⋅ s 2 / ft 2 m B = w B g = 15 lb 32.2 ft/s 2 = 0.46584 lb ⋅ s 2 / ft 2 Show the conservation of momentum of the system as shown in Figure 1. Refer to Figure 1. Consider both the block A and B are at rest initially. The self-weight and normal reaction on the block B are denoted by S w and S N . Consider the velocity attained by the block A and B after cutting of cord, are denoted by v A and v B . Equate the horizontal force on system as follows: 0 + 0 = m A v A − m B v B m B v B = m A v A v B = m A m B v A Substitute 0.15528 lb ⋅ s 2 / ft 2 for m A and 0.46584 lb ⋅ s 2 / ft 2 for m B . v B = 0.15528 0.46584 v A = 1 3 v A (1) Show the conservation of energy of system as shown in Figure 2. Refer to Figure 2. Consider the position 1. Calculate the potential energy ( V 1 e ) of the spring as follows: ( V 1 e ) = 1 2 k e 2 Substitute 72 lb/in . for k and 3 in . for e . ( V 1 e ) = 1 2 × 72 × 3 2 = 324 lb ⋅ in . × ( 1 lb ⋅ ft 12 lb ⋅ in . ) = 27 lb ⋅ ft The potential energy ( V 1 g ) of block A and block B together is 0. The kinetic energy ( T 1 ) of the system in position 1 is 0 as the blocks are at rest. Consider the position 2. The potential energy ( V 2 e ) of the spring is 0

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 14, Problem 14.110RP

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Find the velocity of the blocks A and B just after A leaves B.

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Chapter 14, Problem 14.109RP

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