Asked by Nathan
(ZYX)If the coefficient of friction is 0.85 for a 2500 kg car against asphalt, then what applied horizontal friction force would be necessary to decelerate the car at 6.0 m/s^2? Also, what would be the stopping distance if the vehicles original speed is 25 m/s?
Answers
Answered by
Anonymous
Odd wording. You give the friction force (mu*mg) then ask about an applied "friction" force.
Applied frorce+ friction= ma
applied force+mg*mu=ma
applied force=m(-6-9.8*.85)
now for stopping distance:
vf^2=vi^2+2 a d
0=25^2+2(-6.0)(d)
solve for d.
Applied frorce+ friction= ma
applied force+mg*mu=ma
applied force=m(-6-9.8*.85)
now for stopping distance:
vf^2=vi^2+2 a d
0=25^2+2(-6.0)(d)
solve for d.
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