College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 360: 6

Answer

(a) A force of $415~N$ is required to lift the lid. (b) Since the atmospheric pressure outside the box is lower than the pressure inside the box, the lid would not stay on the box. Instead the lid would get popped off the box from the pressure difference.

Work Step by Step

\sqrt x(a) We can find the pressure difference between the outside and inside of the box: $P_{atm}-P_{in} = 1.013\times 10^5~Pa - 0.80\times 10^5~Pa = 2.13\times 10^4~Pa$ We can find the force pushing down on the lid from the pressure difference: $F = P~A = (2.13\times 10^4~N/m^2)(0.15~m)(013~m) = 415~N$ A force of $415~N$ is required to lift the lid. (b) In Denver, the atmospheric pressure is $67.5~kPa$ which is $0.675\times 10^5~Pa$ Since the atmospheric pressure outside the box is lower than the pressure inside the box, the lid would not stay on the box. Instead the lid would get popped off the box from the pressure difference.
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