56. (a) Using the kinematic equation given in Table 2-1, the deceleration of the car is 222020(35 m/s)2 (107 m)vvada=+¡which gives 25.72 m/s .a=−Thus, the force of friction required to stop by car is 23| |(1400 kg)(5.72 m/s )8.0 10 N.fma==≈×(b) The maximum possible static friction is ,max(0.50)(1400 kg)(9.80 m/s )6.9 10 N.ssgμ≈×(c) If 0.40k=, then kkfmg=and the deceleration is
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.