A 190-kg hockey player sliding across the ice comes to a stopin 11.5m. A) If the

Question

A 190-kg hockey player sliding across the ice comes to a stopin 11.5m. A) If the coefficient of kinetic friction between the theplayer and the ice is 0.500,what is the frictional forceacting on the player? B)How fast was the player going when he fell? c)suppose that the hockey player skated stright into the wallat 5.00m/s and it took 0.150s to stop.(That is the wall applied anaverage stopping force for 0.150s.) How much work was done by thewall in stopping the player? A 190-kg hockey player sliding across the ice comes to a stopin 11.5m. A) If the coefficient of kinetic friction between the theplayer and the ice is 0.500,what is the frictional forceacting on the player? B)How fast was the player going when he fell? c)suppose that the hockey player skated stright into the wallat 5.00m/s and it took 0.150s to stop.(That is the wall applied anaverage stopping force for 0.150s.) How much work was done by thewall in stopping the player?

Explanation / Answer

a ) Mass of the hockey player is m = 190 kg       coefficient of kineticfriction is k   = 0.500            Frictional force Ffr   = -k * FN                                              = 0.500 * 190 kg * 9.81 m/s2                                            = - 931.9 N               acceleration a = F / m                                     = 931.9 N / 190 kg                                    = - 4.9 m/s2   b )   Using Kinematic Equation               v2 - v02 = 2 as             -v02   = (5.00 m/s )2   - 4.9 m/s2 * 0.150s                      v0 = 10.6 m/s c )   Work done W   = F* d                                  = m a * d                                 = 190 kg * - 4.9 m/s2 * 11.5 m                                  = - 10.7 *103 J

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