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At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.

User Sergei R
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1 Answer

1 vote

Answer:

force = 11.33
kg-m/s^(2)

Step-by-step explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is


P_(1)=mU


P_(1)= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion


F=(\Delta P)/(\Delta t)


F = (P_(1)-P_(2))/(T)

Kgm/s^2


F = (69.7-0)/(6.15)= 11.33[tex]kg-m/s^(2)[/tex]

User Sinisa Drpa
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