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Physics 12 Online
OpenStudy (anonymous):

A dog (mass 13.3 kg) sits on a sled (mass 5.5 kg) that is being pushed with a force of 33 N on a flat snowy surface. If the acceleration of both is 1.4 m/s2, what is the coefficient of kinetic friction for the sled sliding on the snow?

OpenStudy (anonymous):

Sorry. I really am bad at this stuff

OpenStudy (anonymous):

Okay are they wanting us to combine the masses or keep them separate?

OpenStudy (photon336):

i would combine them

OpenStudy (photon336):

remember the dog is on the sled both are moving at the same acceleration

OpenStudy (anonymous):

m= 18.8 a = 1.4 m/s^2 Fpush = 33N

OpenStudy (photon336):

|dw:1447122487797:dw|

OpenStudy (photon336):

always make a diagram too

OpenStudy (photon336):

does this make sense to you ? \[F_{applied} - F_{kinetic friction} = Ma_{net} \]

OpenStudy (photon336):

net acceleration is 1.4 m/s^2

OpenStudy (anonymous):

Why is it static friction if the sled is moving?

OpenStudy (photon336):

well there is no more static friction because that was overcome already

OpenStudy (photon336):

but still when you move there's still friction that's opposing the direction of your force and that's called kinetic friction

OpenStudy (anonymous):

Okay so it's 33N-Fkinetic = (18.8 kg)(1.4m/s^2)

OpenStudy (anonymous):

33N-Fkinetic = 26.32N

OpenStudy (anonymous):

-Fkinetic= -6.68

OpenStudy (anonymous):

-1(-Fkintetic= -6.68) = Fkinetic = 6.68N

OpenStudy (anonymous):

Is that it?

OpenStudy (photon336):

yep but they asked you for something else

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