anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ok so the part that confuses me the most is "constant velocity" meaning that there is inertia at work.
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
I am not too sure if I can just use the given values to go with the kinetic friction....Any ideas people?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@nincompoop
anonymous
  • anonymous
The constant velocity part is just another way of saying it's in equilibrium. |dw:1439385917192:dw|
anonymous
  • anonymous
I bet the coefficient of friction in this case must be 0.5
anonymous
  • anonymous
As 80N are required to move an object that weights 160kg
anonymous
  • anonymous
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 \[f=-\mu.N\]\[\mu=-\frac{f}{N}\]\[\mu=-\frac{-80}{160 \times 9.8}=\frac{80}{160 \times 9.8}\]
anonymous
  • anonymous
no. the frictional force must balance the applied force for equilibrium. \[mg \mu_k=80~ N\]
anonymous
  • anonymous
Looks like the second explanation is right.
anonymous
  • anonymous
It is. they're the same. I was replying to the 0.5 response and we posted at about the same time.
anonymous
  • anonymous
So according to the calculation kinetic friction is revealed as 0.05
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.