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?
Sorry. I really am bad at this stuff
Okay are they wanting us to combine the masses or keep them separate?
i would combine them
remember the dog is on the sled both are moving at the same acceleration
m= 18.8 a = 1.4 m/s^2 Fpush = 33N
|dw:1447122487797:dw|
always make a diagram too
does this make sense to you ? \[F_{applied} - F_{kinetic friction} = Ma_{net} \]
net acceleration is 1.4 m/s^2
Why is it static friction if the sled is moving?
well there is no more static friction because that was overcome already
but still when you move there's still friction that's opposing the direction of your force and that's called kinetic friction
Okay so it's 33N-Fkinetic = (18.8 kg)(1.4m/s^2)
33N-Fkinetic = 26.32N
-Fkinetic= -6.68
-1(-Fkintetic= -6.68) = Fkinetic = 6.68N
Is that it?
yep but they asked you for something else
Join our real-time social learning platform and learn together with your friends!