Question #2cc01

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Answer 1 · 2016-12-26T00:23:31

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

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.

$K_i=1/2mv_i^2 = 1/2 (2268)(20.0^2)= 453 600 J$

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

$DeltaK + FDeltad = 0$

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

$-453 600 + (700)Deltad=0$

$Deltad = 453 600/700 = 648 m$

c) To find acceleration, use an equation of motion

$v_f^2 =v_i^2+2aDeltad$

$0 = 20^2 +2a(648)$

$a=-(20^2)/(2xx648) = -0.309 m/s^2$