A 66.0 kg rider sitting on a 8.5 kg bike is riding along at 8.7 m/s in the posit

Question

A 66.0 kg rider sitting on a 8.5 kg bike is riding along at 8.7 m/s in the positive direction. The rider drags a foot on the ground and slows down to 5.1 m/s still in the positive direction. What is the change in momentum of the rider and bike? -2.68×102 kg*m/s

What is the impulse delivered by the ground to the rider's foot? -2.68×102 N*s

What force is acting on the bike and rider if slowing down took 14.3 seconds? -1.88×101 N

And, how far did the bike and rider travel during these 14.3 seconds?

I cannot figure out the last question

Explanation / Answer

a) del(P) = m*(V2-V1) = (66+8.5)*(5.1-8.7) = -268.2 kg*m/s

b) I = del(P) =   -268.2 kg*m/s

c) del(P) = F*t

F = ( -268.2 kg*m/s ) / (14.3) = -18.75 N

d) work done = change in KE = (1/2)*(66+8.5)*(5.1^2 -8.7^2) = -1850.6 J

W = F*d

d =(1850.6/18.75) = 98.7 m

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