Question

Question #2cc01

Answer 1

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`