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PhysicsCollege Physics11th Edition

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Chapter 4, Problem 17P
BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Solutions

Chapter
Section
Problem 1CQ
Problem 2CQ
Problem 3CQ
Problem 4CQ
Problem 5CQ
Problem 6CQ
Problem 7CQ
Problem 8CQ
Problem 9CQ
Problem 10CQ
Problem 11CQ
Problem 12CQ
Problem 13CQ
Problem 14CQ
Problem 15CQ
Problem 16CQ
Problem 17CQ
Problem 18CQ
Problem 19CQ
Problem 20CQ
Problem 21CQ
Problem 22CQ
Problem 23CQ
Problem 24CQ
Problem 1P
Problem 2P
Problem 3P
Problem 4P
Problem 5P
Problem 6P
Problem 7P
Problem 8P
Problem 9P
Problem 10P
Problem 11P
Problem 12P
Problem 13P
Problem 14P
Problem 15P
Problem 16P
Problem 17P
Problem 18P
Problem 19P
Problem 20P
Problem 21P
Problem 22P
Problem 23P
Problem 24P
Problem 25P
Problem 26P
Problem 27P
Problem 28P
Problem 29P
Problem 30P
Problem 31P
Problem 32P
Problem 33P
Problem 34P
Problem 35P
Problem 36P
Problem 37P
Problem 38P
Problem 39P
Problem 40P
Problem 41P
Problem 42P
Problem 43P
Problem 44P
Problem 45P
Problem 46P
Problem 47P
Problem 48P
Problem 49P
Problem 50P
Problem 51P
Problem 52P
Problem 53P
Problem 54P
Problem 55P
Problem 56P
Problem 57P
Problem 58P
Problem 59P
Problem 60P
Problem 61P
Problem 62P
Problem 63P
Problem 64P
Problem 65P
Problem 66P
Problem 67P
Problem 68P
Problem 69P
Problem 70P
Problem 71P
Problem 72AP
Problem 73AP
Problem 74AP
Problem 75AP
Problem 76AP
Problem 77AP
Problem 78AP
Problem 79AP
Problem 80AP
Problem 81AP
Problem 82AP
Problem 83AP
Problem 84AP
Problem 85AP
Problem 86AP
Problem 87AP
Problem 88AP
Problem 89AP
Problem 90AP
Problem 91AP
Problem 92AP
BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A force of 30.0 N is applied in the positive x-direction to a block of mass 8.00 kg, at rest on a frictionless surface. (a) What is the block’s acceleration? (b) How fast is it going after 6.00 s?

(a)

To determine
The acceleration of the block.

Explanation

Given info: The mass of the block is 8.00 kg. The force applied on the block is 30 N.

From Newton’s second law, the formula for acceleration is,

a=Fm

  • m is the mass of the block.
  • F is the force applied on the block.

Conclusion:

Substitute 8.00 kg for m and 30 N for F in the above expression to get N

(b)

To determine
The speed of the block after 6 seconds.

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Chapter 4 Solutions

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Sect-4.2 P-4.1QQSect-4.2 P-4.2QQSect-4.2 P-4.3QQSect-4.2 P-4.4QQSect-4.4 P-4.5QQSect-4.4 P-4.6QQSect-4.4 P-4.7QQSect-4.6 P-4.8QQSect-4.6 P-4.9QQCh-4 P-1CQ
Ch-4 P-2CQCh-4 P-3CQCh-4 P-4CQCh-4 P-5CQCh-4 P-6CQCh-4 P-7CQCh-4 P-8CQCh-4 P-9CQCh-4 P-10CQCh-4 P-11CQCh-4 P-12CQCh-4 P-13CQCh-4 P-14CQCh-4 P-15CQCh-4 P-16CQCh-4 P-17CQCh-4 P-18CQCh-4 P-19CQCh-4 P-20CQCh-4 P-21CQCh-4 P-22CQCh-4 P-23CQCh-4 P-24CQCh-4 P-1PCh-4 P-2PCh-4 P-3PCh-4 P-4PCh-4 P-5PCh-4 P-6PCh-4 P-7PCh-4 P-8PCh-4 P-9PCh-4 P-10PCh-4 P-11PCh-4 P-12PCh-4 P-13PCh-4 P-14PCh-4 P-15PCh-4 P-16PCh-4 P-17PCh-4 P-18PCh-4 P-19PCh-4 P-20PCh-4 P-21PCh-4 P-22PCh-4 P-23PCh-4 P-24PCh-4 P-25PCh-4 P-26PCh-4 P-27PCh-4 P-28PCh-4 P-29PCh-4 P-30PCh-4 P-31PCh-4 P-32PCh-4 P-33PCh-4 P-34PCh-4 P-35PCh-4 P-36PCh-4 P-37PCh-4 P-38PCh-4 P-39PCh-4 P-40PCh-4 P-41PCh-4 P-42PCh-4 P-43PCh-4 P-44PCh-4 P-45PCh-4 P-46PCh-4 P-47PCh-4 P-48PCh-4 P-49PCh-4 P-50PCh-4 P-51PCh-4 P-52PCh-4 P-53PCh-4 P-54PCh-4 P-55PCh-4 P-56PCh-4 P-57PCh-4 P-58PCh-4 P-59PCh-4 P-60PCh-4 P-61PCh-4 P-62PCh-4 P-63PCh-4 P-64PCh-4 P-65PCh-4 P-66PCh-4 P-67PCh-4 P-68PCh-4 P-69PCh-4 P-70PCh-4 P-71PCh-4 P-72APCh-4 P-73APCh-4 P-74APCh-4 P-75APCh-4 P-76APCh-4 P-77APCh-4 P-78APCh-4 P-79APCh-4 P-80APCh-4 P-81APCh-4 P-82APCh-4 P-83APCh-4 P-84APCh-4 P-85APCh-4 P-86APCh-4 P-87APCh-4 P-88APCh-4 P-89APCh-4 P-90APCh-4 P-91APCh-4 P-92AP

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