## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 4 - Newton's Laws of Motion - Problems - Exercises - Page 127: 4.49

#### Answer

(a) The mass of box B is 4.34 kg. (b) The mass of box A is 5.30 kg.

#### Work Step by Step

(a) We can find the downward acceleration of the boxes. $y = \frac{1}{2}at^2$ $a = \frac{2y}{t^2} = \frac{(2)(12.0~m)}{(4.00~s)^2}$ $a = 1.5~m/s^2$ We can use a force equation to find the mass of box B. $\sum F = ma$ $mg - T = ma$ $m(g-a) = T$ $m = \frac{T}{g-a} = \frac{36.0~N}{(9.80~m/s^2)-(1.5~m/s^2)}$ $m = 4.34~kg$ The mass of box B is 4.34 kg. (b) Since the boxes are tied together, the downward acceleration of box A is also $a = 1.5~m/s^2$. We can use a force equation to find the mass of box A. $\sum F = ma$ $mg + T - F = ma$ $m(g-a) = F-T$ $m = \frac{F-T}{g-a} = \frac{(80.0~N)-(36.0~N)}{(9.80~m/s^2)-(1.5~m/s^2)}$ $m = 5.30~kg$ The mass of box A is 5.30 kg.

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