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. Resolve the forces along t − axis to determine the tangential acceleration ( a t ) . ∑ F t = m a t (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 ) : When the box get stopped. The kinetic energy becomes zero. T B = 0 Let, y is the maximum displacement of box. Calculate the potential energy ( V B ) . Substitute 5 kg for m , 9.81 m/s 2 for g and y for h in Equation (II). V B = 5 kg ( 9.81 m/s 2 ) ( y ) = 49.05 y Calculate the maximum distance traveled by the box. Substitute 250 J for T A , 0 for V A , 0 for T B and 49.05 y for V B in Equation (V). 250 J + 0 = 0 + 49.05 y y = 250 49.05 y = 5.0968 m Thus, the maximum distance traveled by the box is 5

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

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…

Videos

Question

Chapter 14.5, Problem 91P

To determine

The maximum distance traveled by the box before it comes to stop, normal reaction and acceleration.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.5, Problem 90P

Chapter 14.5, Problem 92P

Students have asked these similar questions

A pump delivering 230 lps of water at 30C has a 300-mm diameter suction pipe and a 254-mm diameter discharge pipe as shown in the figure. The suction pipe is 3.5 m long and the discharge pipe is 23 m long, both pipe's materials are cast iron. The water is delivered 16m above the intake water level. Considering head losses in fittings, valves, and major head loss. a) Find the total dynamic head which the pump must supply. b)It the pump mechanical efficiency is 68%, and the motor efficiency is 90%, determine the power rating of the motor in hp.

The tensile 0.2 percent offset yield strength of AISI 1137 cold-drawn steel bars up to 1 inch in diameter from 2 mills and 25 heats is reported as follows: Sy 93 95 101 f 97 99 107 109 111 19 25 38 17 12 10 5 4 103 105 4 2 where Sy is the class midpoint in kpsi and fis the number in each class. Presuming the distribution is normal, determine the yield strength exceeded by 99.0% of the population. The yield strength exceeded by 99.0% of the population is kpsi.

Solve this problem and show all of the work

Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 14.3 - Determine the work of the force when it displaces...Ch. 14.3 - Determine the kinetic energy of the 10-kg block.Ch. 14.3 - Prob. 1FP Ch. 14.3 - If the motor exerts a constant force of 300 N on...Ch. 14.3 - If the motor exerts a force of F = (600 + 2s2) N...Ch. 14.3 - The 1.8-Mg dragster is traveling at 125 m/s when...Ch. 14.3 - When s = 0.5 m, the spring is unstretched and the...Ch. 14.3 - The 5-lb collar is pulled by a cord that passes...Ch. 14.3 - The 20-kg crate is subjected to a force having a...Ch. 14.3 - Prob. 2P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The maximum distance traveled by the box before it comes to stop, normal reaction and acceleration.

Solution Summary: The author explains how to determine the maximum distance traveled by the box before it comes to stop, normal reaction and acceleration.

Concept explainers

Videos

The maximum distance traveled by the box before it comes to stop, normal reaction and acceleration.

Want to see the full answer?

Chapter 14 Solutions