The sled dog must produce a horizontal force of 80 N in order to keep the 160 kg sled moving across level ice at a constant velocity. What is the coefficient of kinetic friction between the sled and the ice?

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The sled dog must produce a horizontal force of 80 N in order to keep the 160 kg sled moving across level ice at a constant velocity. What is the coefficient of kinetic friction between the sled and the ice?

Mathematics
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Ok so the part that confuses me the most is "constant velocity" meaning that there is inertia at work.
I am not too sure if I can just use the given values to go with the kinetic friction....Any ideas people?

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Other answers:

The constant velocity part is just another way of saying it's in equilibrium. |dw:1439385917192:dw|
I bet the coefficient of friction in this case must be 0.5
As 80N are required to move an object that weights 160kg
By constant velocity they mean that acceleration is 0, and therefore there is no resultant force on the sled, then the force applied by the dog is just enough to balance friction \[\implies f=-80N\] taking the direction of motion of sled as positive direction, then friction will work in opposite direction Now we know that \[f=-\mu.N\]\[\mu=-\frac{f}{N}\]\[\mu=-\frac{-80}{160 \times 9.8}=\frac{80}{160 \times 9.8}\]
no. the frictional force must balance the applied force for equilibrium. \[mg \mu_k=80~ N\]
Looks like the second explanation is right.
It is. they're the same. I was replying to the 0.5 response and we posted at about the same time.
So according to the calculation kinetic friction is revealed as 0.05
Well that seems pretty low but I guess that's how much kinetic friction is required for the dog to just apply 80N force and have the sled moving ahead.
The low coefficient of friction makes sense because 1) the surface is ice and 2) the coefficient of kinetic friction is generally less than the coefficient of static friction because it takes more force to make something move than it does to keep it moving.

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