AP Physics Help (Ski-er on Slope)

[weiseguy]

1. http://sites.google.com/site/startngovr/
2. $$\Sigma$$F = $$\Sigma$$m*a
F$$_{k}$$=$$\mu$$*N
X$$_{f}$$ = 1/2 at$$^{2}$$
V$$_{f}$$ = at

For problem 1 we think the answers are as followed:
$$\mu_{s}$$ does not = 0.2
a = 5.32m/s$$^{2}$$

For problem 2 we have no idea what to do. We have too many unknowns. We believe it is not enough info.

For problem 3, we didn't finish, but with what we did have we wound up with a negative velocity, which isn't the case correct? We solved for acceleration to be -150m/s$$^{2}$$. We then found time to = .023094 seconds. Using that information, we found velocity to be (-150)(.023094) = a negative value. This cannot be true can it?

 

[Astronuc]

For problem 2 we have no idea what to do. We have too many unknowns. We believe it is not enough info.

If that is the sliding block tied to a hanging block with a line over the pulley, then one has enough information.

Part of the problem is to determine if μ_s is sufficient to maintain a static system, so one has to determine if the tension in the rope, T, is < or > than the friction force for μ_s = 0.2.

If the tension is > than the static friction force, then the block must be sliding, and so one has to determine the acceleration of both blocks, in which case one has to consider the mass of both blocks, which accelerate together.

 

[weiseguy]

If that is the sliding block tied to a hanging block with a line over the pulley, then one has enough information.

Part of the problem is to determine if μ_s is sufficient to maintain a static system, so one has to determine if the tension in the rope, T, is < or > than the friction force for μ_s = 0.2.

If the tension is > than the static friction force, then the block must be sliding, and so one has to determine the acceleration of both blocks, in which case one has to consider the mass of both blocks, which accelerate together.

Yes we know that. We actually have that one done. Problem 2 is the skier on the slope.

 

[Astronuc]

One can resolve the weight of the skier and the friction, which is proportional to the weight, so once can determine the downhill (parallel to the incline) acceleration which is proportional to g.

Resolve the weight into normal force on incline and forces parallel to incline (friction up the incline and accelerating force down the incline).

Here's a nice reference - http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html#c2

 

[weiseguy]

One can resolve the weight of the skier and the friction, which is proportional to the weight, so once can determine the downhill (parallel to the incline) acceleration which is proportional to g.

Resolve the weight into normal force on incline and forces parallel to incline (friction up the incline and accelerating force down the incline).

Here's a nice reference - http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html#c2

Ummm...what?

I don't understand how to deduce the weight/mass of the skier still.

 

[weiseguy]

Bump, its due tomorrow!

 

[Astronuc]

Ummm...what?

I don't understand how to deduce the weight/mass of the skier still.

Weight is a force, and W = mg, where g is the acceleration (also a vector). Gravity always points down, but on an incline, the weight vector can be resolved into two components, one normal to the surface of the incline and the other parallel.

 

[weiseguy]

yes, I understand finding the 2 components of gravity in both the horizontal and vertical, but I do not have a mass to multiply the acceleration due to gravity by.

mg means mass*9.8 :P