Explanation Given: The mass of the block is 20 kg . The velocity of the block is v = 5 m/s . The coefficient of friction is μ k = 0.2 . Draw the free body diagram of blocks as shown in Figure (1). Write the formula for power ( P ) . P = F v (I) Here, F is the force and v is the velocity. Write the formula for friction force ( F f ) . F f = μ k N (II) Here, μ k is the coefficient of kinematic friction and N is the normal reaction force. Refer Figure (1). Resolve the forces along y − axis to calculate the normal reaction force ( N ) . ∑ F y = m a y Here, the acceleration along y − a x i s is zero because there is no displacement. Substitute 0 for a y in Equation ∑ F y = m a y . ∑ F y = m ( 0 ) ∑ F y = 0 (III) Refer Figure (1). Resolve the forces along x − axis to calculate the force ( F ) . ∑ F x = m a x Here, the acceleration along x − axis is also zero because the block moves with constant velocity. Substitute 0 for a x in Equation ∑ F x = m a x . ∑ F x = m ( 0 ) ∑ F x = 0 (IV) Conclusion: Calculate the normal reaction force ( N )

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.4, Problem 10FP

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The power of the force (F) .

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The power of the force ( F ) .

Solution Summary: The author explains how to determine the power of the force (F) on the block at the constant velocity of 5m/s.

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Chapter 14 Solutions