Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 113: 6

Answer

The projectile requires 0.041 s to reach that speed.

Work Step by Step

1. Calculate the acceleration of the projectile, using the net force and the mass: $F = m\times a$ $4.9 \times 10^5N = 5.0kg \times a$ Divide both sides by $5.0kg$. $\frac{4.9 \times 10^5N}{5.0kg} = a$ $9.8 \times 10^4$ $m/s^2 = a$ 2. Find the time that is required to get $4.0 \times 10^3$ $m/s$: $V = V_0 + at$ - Since the projectile starts with no speed (rest), $V_0 = 0 m/s$ $V = at$ $4.0 \times 10^3 m/s = 9.8 \times 10^4 m/s^2 \times t$ $\frac{4.0 \times 10^3m/s}{9.8 \times 10^4m/s^2} = t$ $t = 0.041$ $s$
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