A 66.0 kg rider sitting on a 8.0 kg bike is riding along at 9.3 m/s in the posit

Question

A 66.0 kg rider sitting on a 8.0 kg bike is riding along at 9.3 m/s in the positive direction. The rider drags a foot on the ground and slows down to 5.4 m/s still in the positive direction.

What is the change in momentum of the rider and bike? = -2.89x10^2 kg*m/s
What is the impulse delivered by the ground to the rider's foot? -2.89×102 N*s

What force is acting on the bike and rider if slowing down took 15.2 seconds? -1.90×101 N

And, how far did the bike and rider travel during these 15.2 seconds???

Explanation / Answer

a)

Momentum is given by

P=mV

Change in momentum bo rider and bike is

dP =Pfinal-Pinital =(66+8)*5.4-(66+8)*9.3

dP=-288.6 Kg-m/s = -2.89*102 Kg-m/s

b)

Impulse

dP ==-288.6 Kg-m/s = -2.89*102 Kg-m/s

c)

Force

F=dP/dt =-2.89*102/15.2

F=-1.9*101 N

d)

Average velocity

Vavg =(9.3+5.4)/2=7.35 m/s

Distance travelled

X=Vavgt =7.35*15.2

X=111.72 m

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