Posted by Jason on Wednesday, May 25, 2011 at 9:53pm.
The friction force on the tires remains -4.40 m/s^2*M if tires are on the verge of slipping. Gravity will apply an additional retarding force of -M g sin 12 = 2.04 m/s^2*M
The resultant deceleration is
(-4.40 - 2.04) M/M
= -6.44 m/s^2
(mass M cancels out)
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