Explanation Given: The mass of the box is 5 kg . The speed of the box at point A is v A = 10 m/s . The equation of path is x 1 / 2 + y 1 / 2 = 3 . Draw the free body diagram of boxas shown in Figure (1). The datum is set at position A . Write the formula for kinetic energy ( T ) . T = 1 2 m v 2 (I) Here, m is the mass and v is the velocity. Refer Figure (1). Write the formula for gravitational potential energy ( V g ) . V g = m g h (II) Here, m is the mass of the box, g is the acceleration due to gravity and h is the displacement. Here, the gravitational potential energy ( V g ) is equal to the total potential energy. V = V g Write the formula for conservation of energy. T A + V A = T B + V B (III) Here, T A , T B are the initial and final kinetic energy and V A , V B are the initial and final potential energy. Resolve the forces along n − axis to determine the normal reaction force ( N ) . ∑ F n = m a n (IV) Here, m is the mass of box and a n is the acceleration along axis − n . When the box traveling along the path of x 1 / 2 + y 1 / 2 = 3 . The acceleration of the path is a n = v 2 ρ . Substitute v 2 ρ for a n in Equation (VI). ∑ F n = m ( v 2 ρ ) (V) Here ρ is the radius of the curvature. ρ = [ 1 + ( d y d x ) 2 ] 3 2 | d 2 y d x 2 | (VI) Conclusion: At position ( A ) : Calculate the initial kinetic energy the box. Substitute 5 kg for m and 10 m/s for v in Equation (I). T A = 1 2 ( 5 kg ) ( 10 m/s ) 2 = 250 J Refer Figure (1). When the box is at the datum. The potential energy of the box is V A = 0 . At position ( B ) : Refer Figure (1). Calculate displacement of the box at the point B . y = x Substitute x for y in the Equation x 1 / 2 + y 1 / 2 = 3 . x 1 / 2 + x 1 / 2 = 3 2 x 1 / 2 = 3 x 1 / 2 = 3 2 x = ( 3 2 ) 2 x = 9 4 m Similarly, y = 9 4 m . Calculate the final kinetic energy. Substitute 5 kg for m and v B for v in Equation (I). T B = 1 2 ( 5 kg ) ( v B ) 2 = 2.5 v B 2 Calculate the potential energy ( V B ) . Substitute 5 kg for m , 9.81 m/s 2 for g and 9 4 m for h in Equation (II). V B = 5 kg ( 9.81 m/s 2 ) ( 9 4 m ) = 110.3625 J Calculate the speed of the box at point B . Substitute 250 J for T A , 0 for V A , 2.5 v B 2 for T B and 110.3625 J for V B in Equation (V). 250 J + 0 = 2.5 v B 2 + 110.3625 J 2 .5 v B 2 = 250 − 110.3625 v B 2 = 139

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its enviro…

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Chapter 14.5, Problem 90P

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