You decide to transfer to the University of Alaska at Fairbanks because a friend

Question

You decide to transfer to the University of Alaska at Fairbanks because a friend told you that they have a fantastic, week-long Winter Carnival. One event is a dogsled race in which teams of students replace the dogs. The students wear cleats in order to get traction on the ice. You arrange for a team of 5 other students and yourself as the pullers and one more student to ride on the sled as the musher. You understand that traditionally the pullers get revenge on the musher at a party at the end of the race. The pullers on your team each have a mass of 80 kg, the musher has a mass of 60 kg and the sled, which has been loaded with sandbags, has a mass, including the musher, of 200 kg. There is negligible friction between the sled and the ice. The pullers can each generate a maximum force of 220.0 N on the harness, which makes an angle of 20.6 degrees to the horizontal at the point at which it attaches to the sled. The pullers can also run at a maximum speed of 4.50 m/s.

If the sled begins at rest, how long does it take before it reaches its maximum speed of 4.50 m/s?


The race course is 500 m long. How long does it take for your team to reach the finish, assuming that you start at rest?

Explanation / Answer

there are 6 pullersso total mass of pullers is 6*80=480kg

mass of musher=60kg

xtra weight=200kg

total mass which is to be pulled M=200+60+480=740kg

forch applied by each man is 220N at angle of 20.6 with horizontal

so force responsible for forward motion is 220cos20.6N

since there are 6 pullers so total force F=6*220*cos(20.6)N

since friction is negligible so f=0

F=Ma

a=F/M =1320cos20.6/740

a=1.67m/s2

inital u=0m/sec

final v=4.5m/sec

v-u=at

4.5=1.67t

=>t=4.5/1.67

=>t=2.695sec

since the max value of velocity is 4.5m/sec so we can assume that they move with constant velocity

after they reach 4.5m/sec

so distance travelled in accelerating upto 4.5m/sec in t=2.695 is

s=ut+.5at^2

s=0+.5*1.67*2.695^2

=6.063

so now they travel 500-6.063 with 4.5m/sec

so d=vt

500-6.063=4.5t

t=493.937/4.5

t=109.764sec

total time required to travel 500m is 109.764+2.695

total time=112.45882sec=1.8743min

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