G 26. The stopping force F and the path of the passenger are horizontal. Our +x axis is in the direction of the passenger’s motion, so that the passenger’s acceleration (‘‘deceleration” ) is negative-valued and the stopping force is in the –x direction: G F = − F ˆi . Using Eq. 2-16 with v0 = (53 km/h)(1000 m/km)/(3600 s/h) = 14.7 m/s and v = 0, the acceleration is found to be v 2 = v02 + 2aΔx a = −
v02 (14.7 m/s) 2 =− = −167 m/s 2 . 2Δx 2 ( 0.65 m )
Assuming there are no significant horizontal forces other than the stopping force, Eq. 5-1 leads to G G F = ma − F = 41 kg −167 m s2
b
which results in F = 6.8 × 103 N.
gc
h
