30. The stopping force GFand the path of the car are horizontal. Thus, the weight of the car contributes only (via Eq. 5-12) to information about its mass (m= W/g= 1327 kg). Our +xaxis is in the direction of the car’s velocity, so that its acceleration (‘‘deceleration”) is negative-valued and the stopping force is in the –xdirection: ˆiFF=−G.(a) We use Eq. 2-16 and SI units (noting that v= 0 and v0= 40(1000/3600) = 11.1 m/s). ()222200(11.1 m/s)215mvvvaxax=+Δ¡= −Δwhich yields a= – 4.12 m/s2. Assuming there are no significant horizontal forces other than the stopping force, Eq. 5-1 leads to GGFmaF=¡−=−1327412kgm s2
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