G 28. The stopping force F and the path of the toothpick are horizontal. Our +x axis is in the direction of the toothpick’s motion, so that the toothpick’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 = 220 m/s and v = 0, the acceleration is found to be
v 2 = v02 + 2aΔx a = −
v02 (220 m/s) 2 =− = −1.61× 106 m/s 2 . 2Δx 2 ( 0.015 m )
Thus, the magnitude of the force exerted by the branch on the toothpick is F = m | a | = (1.3 ×10−4 kg)(1.61×106 m/s 2 ) = 2.1×102 N.
