Question #2cc01

Dec 26, 2016

a) The initial kinetic energy is 453 600 J.
b) The stopping distance is 648 m
c) Acceleration while stopping averages $- 0.309 \frac{m}{s} ^ 2$

Explanation:

a) The initial kinetic energy is a matter of the mass and the initial speed, and does not depend on the way in which the truck came to a halt.

${K}_{i} = \frac{1}{2} m {v}_{i}^{2} = \frac{1}{2} \left(2268\right) \left({20.0}^{2}\right) = 453 600 J$

b) The stopping distance is found by using conservation of energy to change the kinetic energy to frictional heat

$\Delta K + F \Delta d = 0$

Since the final kinetic energy is zero, the change in K is -453 600 J

$- 453 600 + \left(700\right) \Delta d = 0$

$\Delta d = 453 \frac{600}{700} = 648 m$

c) To find acceleration, use an equation of motion

${v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta d$

$0 = {20}^{2} + 2 a \left(648\right)$

$a = - \frac{{20}^{2}}{2 \times 648} = - 0.309 \frac{m}{s} ^ 2$