A 87.2-kg Baseball Player Slides Into Second Base. The Coefficient Of Kinetic Friction Between The Player And?
Tuesday, November 17th, 2009 at
3:52 am
A 87.2-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.502. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.40 s, what is his initial speed?
Tagged with: 87.2kg • Base • Baseball • Coefficient • Friction • Into • Kinetic • Player • Second • Slides
Filed under: boys of summer
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Frictional Force = 428.98 N
Vi = 6.87 m/s
Frictional Force = Sliding Force * μk
= m * g * μk
= 87.2* 9.81* 0.502
= 429.4268N
Net force = Sliding Force – Friction
= 87.2 * 9.81 – 429.4268
= 426 .0052N
Net force = Mass * Acceleration = Mass* ((Vinitial – Vfinal)/Time)
Vinitial = ?
Vfinal = 0
Time = 1.40s
Vinitial = 426 * 1.4/87.2 = 6.839 m/s