A motorcycle racer traveling at 145 km/h loses control in a corner of the track and slides across the concrete surface. The combined mass of the rider and bike is 243 kg. The steel of the motorcycle rubs against the concrete road surface. (a) What is the frictional force between the road and the motorcy- cle and rider? (b) What would be the acceleration of the motorcycle and rider during the wipeout? (c) Assuming there were no barriers to stop the motorcycle and rider, how long would it take the bike and the rider to slow to a stop?
3 answers
you need the coefficient of friction to solve this
Actually, you don't need the coefficent of friction, you need the distance he slid, or the time.
Vf^2=vi^2 + 2ad but a=forcefriction/mass
and if you knew distance d, you could solve for forcefriction
Vf^2=vi^2 + 2ad but a=forcefriction/mass
and if you knew distance d, you could solve for forcefriction
data:
V= 145 km/h x 1000 m/1 km x 1 h/3600 s = 40.28 m/s
M=243 kg
U=0.30 (coefficient of friction from the book)
solution:
a) Weight= m.g (gravity); W=243 kg x 9.80 m/s2
W=2381.4 N
Fnet-W=0 ; Fnet=W
(friction) Fr= u.Fnet
Fr=0.30 x 2381.4 N
Fr=714.4 N
b) Fr-F=0; Fr=F
F=m x a
a= F/m
a= 714.4 N(kg.m/s2) /243 kg
a= -2.9 m/s2 (negative because is deceleration)
c) F=m.v/t
t=m.v/F
t=243 kg x 40.28 m/s / 714 ( kg x m/s2)
t = 13.7 s
V= 145 km/h x 1000 m/1 km x 1 h/3600 s = 40.28 m/s
M=243 kg
U=0.30 (coefficient of friction from the book)
solution:
a) Weight= m.g (gravity); W=243 kg x 9.80 m/s2
W=2381.4 N
Fnet-W=0 ; Fnet=W
(friction) Fr= u.Fnet
Fr=0.30 x 2381.4 N
Fr=714.4 N
b) Fr-F=0; Fr=F
F=m x a
a= F/m
a= 714.4 N(kg.m/s2) /243 kg
a= -2.9 m/s2 (negative because is deceleration)
c) F=m.v/t
t=m.v/F
t=243 kg x 40.28 m/s / 714 ( kg x m/s2)
t = 13.7 s
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