## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 9 - Exercises and Problems - Page 166: 82

#### Answer

a) $J=38,427\ Ns$ b) $192,133 \ N$ c) $2767\ kg$

#### Work Step by Step

a) We know that the impulse equals the integral of the F(t) function: $J = \int_0^.2(at^4+bt^3+ct^2+d)dt$ Plugging in the given constants and taking the integral, we find: $J=38,427\ Ns$ b) The average force equals the impulse divided by the change in time: $F_{avg}=\frac{38,427}{.2}=192,133 \ N$ c) We use Newton's second law: $m=\frac{F}{a} = \frac{192,133}{69.44}=2767\ kg$

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