Question #2cc01

1 Answer
Dec 26, 2016

Answer:

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#

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=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#