Explanation Given: The mass of the block is 10 kg The force acting at B is F = 250 N (Vertically downward). The displacement of block is 1.5 m (Vertically upward). Draw the free body diagram of the block with reference to datum as shown in Figure (1). Refer Figure (1). Write the formula for work done by the force ( F ) on the block. U F = F ( δ S F ) (I) Here, F is the force and δ S F is the change in displacement due to force ( F ) . Write the formula for work done by its weight on the block. U W = m g ( δ S W ) (II) Here, m is the mass, g is acceleration due to gravity and δ S W is the change in displacement due to the force exerted by the weight of the block. Write the formula for kinetic energy ( T ) T = 1 2 m v 2 (III) 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 = T 2 (IV) Here, T 1 is initial kinetic energy, T 2 is final kinetic energy and ∑ U is total work done by all the forces on the block. Conclusion: From Figure (1). With reference to the datum, the total length of the chord is equal to the addition of the chord length ( S W ) and two times the chord length of ( S F ) . S W + 2 S F = l The change in their chord lengths due the action of forces will be. δ S W + 2 δ S F = 0 When, the block displaced by the action of force ( F ) . Substitute − 1.5 for δ S W . − 1.5 + 2 δ S F = 0 δ S F = 1.5 2 δ S F = 0.75 m Calculate the Work done by the force ( F ) . Substitute 250 for F and 0.75 for δ S F in Equation (I)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.3, Problem 8P

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The speed of the block when it moved by 1.5 m .

Solution Summary: The author explains the formula for work done by the force (F) on the block.

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The speed of the block when it moved by 1.5 m .

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