Finding the angle of an inclined plane

[Mitza]

So here's the problem:

A sled slides down a long snow-covered slope that is at an angle θ to the horizontal. Kinetic friction acts on the sled as it slides where the kinetic coefficient of friction between the sled and snow is μk = 0.050. Ignore air-resistance when solving this problem.

If we observe the sled to be traveling at a constant velocity, what is θ? (in degrees)

So far I've drawn a force diagram and done this;
∑F= ma
∑F= 0
mgsinθ - μmgcosθ = 0
mgsinθ = μcosθ
tanθ = μ
arctanμ = θ

I know I've done something wrong because with that working the final angle is 2.9° which seems far too small.

Thanks in advance for any help!

 

[haruspex]

So here's the problem:

A sled slides down a long snow-covered slope that is at an angle θ to the horizontal. Kinetic friction acts on the sled as it slides where the kinetic coefficient of friction between the sled and snow is μk = 0.050. Ignore air-resistance when solving this problem.

If we observe the sled to be traveling at a constant velocity, what is θ? (in degrees)

So far I've drawn a force diagram and done this;
∑F= ma
∑F= 0
mgsinθ - μmgcosθ = 0
mgsinθ = μcosθ
tanθ = μ
arctanμ = θ

I know I've done something wrong because with that working the final angle is 2.7° which seems far too small.

Thanks in advance for any help!

Your answer is correct.
It probably seems too shallow because the static coefficient is a lot higher, more like 0.1-0.15. So you have to assume the sled was given a nudge to get it started.

 

[Mitza]

Your answer is correct.
It probably seems too shallow because the static coefficient is a lot higher, more like 0.1-0.15. So you have to assume the sled was given a nudge to get it started.

Okay thank you for your response!