Explanation Given: The graph of force ( F ) varies with the time ( t ) is shown in Figure (1a) and the graph of velocity ( v ) varies with the time ( t ) as shown in Figure (1b). Write the formula for power applied on the machine. P = F v (I) Here, F is the force applied on the machine and v is the velocity. Write the formula for work done. U = ∫ t 1 t 2 P d t (II) Here, P is the power applied on the machine and t 1 , t 2 are the initial and final time. Conclusion: Refer Figure (1a). The force applied on the machine is F = 800 N . Calculate the velocity at the time interval of t = 0 s to 0 .2 s . From Figure (1b). v − 0 t − 0 = 20 − 0 0.3 − 0 v = 20 0.3 t v = 66.67 t Calculate the power applied on the machine at the time interval of t = 0 s to 0 .2 s . Substitute 800 N for F and 66.67 t for v in Equation (I). P 1 = 800 N × ( 66.67 t ) m/s = 53333.33 t W × 1 kW 1000 W = 53.3 t kW Refer Figure (1a). Calculate the force applied on the machine at the time interval of t = 0.2 s to 0 .3 s . From Figure (1a). F − 800 t − 0.2 = 0 − 800 0.3 − 0.2 F − 800 = − 800 0.1 ( t − 0.2 ) F = − 8000 t + 1600 + 800 F = 2400 − 8000 t Calculate the velocity at the time interval of t = 0.2 s to 0 .3 s . From Figure (1b). The velocity is curve is straight line at the period of t = 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.4, Problem 63P

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Themaximum power developed during the time period of t=0.3 s .

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Chapter 14.4, Problem 62P

Chapter 14.4, Problem 64P

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Themaximum power developed during the time period of t = 0.3 s .

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