Physics Q pack #82

[deleted647690]

I was reviewing this problem, and I'm getting confused about what theta is in the equation for W=Fxcostheta

I know it says that theta in this case for the Force of friction and displacement is 180 degrees, but I thought that the angle of the incline corresponded to the angle used in the equation.

If it had asked for the work done by gravity with an increase in the angle of the incline, would it increase? Because the force due to gravity on an incline is

F = mgsin theta, where theta is in fact the angle of the incline.


Also, how does increasing the incline decrease the normal force? Is it because as you increase the incline, the normal force becomes more significantly split into its horizontal and vertical components?
I don't see how increasing the incline would decrease the normal force. I would think it increases the normal force since the normal force is influenced by the angle of the incline.....I'm pretty confused

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[Nugester]

Think of it like this. What happens when the incline is completely vertical? There is no normal force! So you would expect that when the angle gradually increases, the normal force gradually diminishes. Work done by friction is = ukN*d, so as N decreases, so does the work done by friction.

As for the work formula: you are confused because all the prep books when going over angles in inclines and work on a box use the same variable. To keep it clear, give the angle of the incline a different variable, like omega and keep theta in the work formula.
So Fx of the box down the incline = mgsine(omega). Work done by kinetic friction = F*dcos(theta) = ukN*d; here one can see that as omega increases, as F is in the same opposite direction to d, theta = 180 degrees, normal force decreases and work decreases.

 

[deleted647690]

Think of it like this. What happens when the incline is completely vertical? There is no normal force! So you would expect that when the angle gradually increases, the normal force gradually diminishes. Work done by friction is = ukN*d, so as N decreases, so does the work done by friction.

As for the work formula: you are confused because all the prep books when going over angles in inclines and work on a box use the same variable. To keep it clear, give the angle of the incline a different variable, like omega and keep theta in the work formula.
So Fx of the box down the incline = mgsine(omega). Work done by kinetic friction = F*dcos(theta) = ukN*d; here one can see that as omega increases, as F is in the same opposite direction to d, theta = 180 degrees, normal force decreases and work decreases.

Thanks, I think I finally got it. Since the normal force is equivalent to the vertical component of gravity, it is = mg cos (sigma), where sigma is the angle of the incline. Since cos*sigma decreases with increasing value of sigma, the normal force must be decreasing. Thus, the force provided by friction is decreasing as well, and so is the resultant work done.

 

[wannabedoc0209]

The explanation above makes complete sense, but I'm still hung up on one thing...the passage gives the formula for kinetic energy, Mgd where d is the height of the ramp. If the angle of the ramp to the horizontal is increased, wouldn't d increase as well (intuitively)? therefore increasing the work done? I hope this makes sense, I'm getting all confused, and any explanation would be much appreciated!

 

[BerkReviewTeach]

The explanation above makes complete sense, but I'm still hung up on one thing...the passage gives the formula for kinetic energy, Mgd where d is the height of the ramp. If the angle of the ramp to the horizontal is increased, wouldn't d increase as well (intuitively)? therefore increasing the work done? I hope this makes sense, I'm getting all confused, and any explanation would be much appreciated!

This is a very common ambiguity in this type of question. What you need to ask (and answer) is whether the initial height of the object is changing when you move the ramp. If the length of the ramp is remaining content, then yes, the height must increase as the angle with the horizontal ground increases. If, however, the length of the ramp changes and the height of the starting point remains the same, then the potential energy at the start never changes.

This a great example of how you can know everything you need to know about a subject, but it comes down to translating what their question is asking.

 

[wannabedoc0209]

This is a very common ambiguity in this type of question. What you need to ask (and answer) is whether the initial height of the object is changing when you move the ramp. If the length of the ramp is remaining content, then yes, the height must increase as the angle with the horizontal ground increases. If, however, the length of the ramp changes and the height of the starting point remains the same, then the potential energy at the start never changes.

This a great example of how you can know everything you need to know about a subject, but it comes down to translating what their question is asking.

Thank you for the clarification! I think I got it now! But just to double check: In this particular problem, we're assuming that the length of the incline is changing with the angle, therefore keeping the height constant?

 

[BerkReviewTeach]

In Question #82 it ends up being a moot point. As the angle is increased, the magnitude of the normal force decreases, resulting in less work done by friction over a given distance. So, if the ramp distance remains the same (as they allude to in their answer explanation), the work done by friction (W = µk x mg cos (theta) x d) decreases because cos (theta) decreases. If the ramp distance diminishes (to keep the initial height the same), then work done by friction still decreases (by even more now), given that both cos(theta) and d are decreasing.