Explanation Given data: Refer to the Figure P5.51 in the textbook for the velocity graph of the 2.0 kg object at various time instants. Mass of the object ( m ) is 2.0 kg. Formula used: Write the expression for net force of the object in terms of mass and acceleration as follows. F net = m a x (1) Here, m is the mass of the object, and a x is the acceleration of the object in the x -direction (horizontal direction). Write the expression to find the acceleration as follows. a x = v f − v i t f − t i (2) Here, v i is the initial velocity of the object, v f is the final velocity of the object, t i is the initial time, and t f is the final time. Explanation: From the figure P5.51 in the textbook, the acceleration is in three segments. The increasing velocity from 0 to 3 s, constant velocity (zero acceleration segment) from 3 s to 6 s, and the decreasing velocity from 6 s to 8 s. The downward acceleration from 6.0 s to 8 s is considered with negative sign. Calculation of acceleration from 0 s to 3 s: From the figure, the initial velocity is 0 m s , final velocity is 12 m s , initial time is 0 s, and the final time is 3 s. Substitute 0 m s for v i , 12 m s for v f , 0 s for t i , and 3 s for t f in equation (2), a x = 12 m s − 0 m s 3 s − 0 s = 12 m s 3 s = 4 m s 2 As the acceleration of the object from 0 s to 3 s is 4 m s 2 , the acceleration at t = 1 s is 4 m s 2 . Calculation of net force of the object at t = 1 s : Substitute 2.0 kg for m , and 4 m s 2 for a x in equation (1), F net = ( 2.0 kg ) ( 4 m s 2 ) = 8 kg ⋅ m s 2 = 8 N { ∵ 1 kg ⋅ m s 2 = 1 N } Therefore, the net force of the object at t = 1 s is 8 N . Calculation of acceleration from 3 s to 6 s: From the figure, the initial velocity is 12 m s , final velocity is 12 m s , initial time is 3 s, and the final time is 6 s. Substitute 12 m s for v i , 12 m s for v f , 3 s for t i , and 6 s for t f in equation (2), a x = 12 m s − 12 m s 6 s − 3 s = 0 m s 3 s = 0 m s 2 As the acceleration of the object from 3 s to 6 s is 0 m s 2 , the acceleration at t = 4 s is 0 m s 2

Modified Mastering Physics with Pearson eText -- Standalone Access Card -- for College Physics: A Strategic Approach (4th Edition)

4th Edition

ISBN: 9780134724744

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 5, Problem 51GP

To determine

To find: The net force acting on the object at t=1 s, at t=4 s, and at t=7 s.

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Chapter 5, Problem 50GP

Chapter 5, Problem 52GP

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Chapter 5 Solutions

Modified Mastering Physics with Pearson eText -- Standalone Access Card -- for College Physics: A Strategic Approach (4th Edition)

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Knowledge Booster

The net force acting on the object at t = 1 s , at t = 4 s , and at t = 7 s .

To find: The net force acting on the object at t=1 s, at t=4 s, and at t=7 s.

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Chapter 5 Solutions