Explanation Given: The weight of the car ( W ) is 4000 lb . The relation between the force and deflection of the barrier is F = [ 90 ( 10 3 ) x 1 2 ] lb . The initial speed of the car is 75 ft/s . Draw the free body diagram of the car as shown in Figure (1). Write the formula to calculate the work done by the force ( F b ) . U F b ( 1 − 2 ) = ∫ s 1 s 2 F d s (I) Here, U F b ( 1 − 2 ) is the work done by the force ( F b ) . Write the formula for kinetic energy ( T ) . T = 1 2 m v 2 (II) Here, m is the mass of the object and v is the velocity of the object. Write the formula for Principle of Work and Energy. T 1 + ∑ U 1 − 2 = T 2 (III) Here, T 1 is initial kinetic energy, T 2 is final kinetic energy and ∑ U 1 − 2 is total work done by all the forces on the car. Refer Figure (1). Horizontal Force components: The horizontal component of force ( F b ) is acting from barrier towards car and does negative work. When the car hits the barrier, the penetration of the car is based on the force exerted by the barrier against the car and the net work done will be product of the force exerted by the barrier and the car’s penetration distance. Vertical Force components: The vertical force components do not create any displace of car hence the work done by these force components are equal to zero. Conclusion: Calculate total work done ( ∑ U 1 − 2 ) . ∑ U 1 − 2 = − U F b ( 1 − 2 ) Calculate the Work done by the force ( F b ) . Substitute 90 ( 10 3 ) x 1 2 for F , 0 for s 1 , x max for s 2 and d x for d s in Equation (I)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The car’s maximum penetration in the barrier.

Solution Summary: The author explains how the car's maximum penetration in the barrier is 3.24ft.

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