Recent content by anne921

Homework Statement A professional skier starts from rest and reaches a speed of 48 m/s on a ski slope angled 22.0° above the horizontal. Using the work-KE theorem and disregarding friction, find the minimum distance along the slope the skier would have to travel in order to reach this...

Thanks!

I am still struggling with this one. I was able to solve it I think... the parallel component of gravity is mgsin30 which is 33.1 N; so the frictional force must balance this out so Fn = Ff/0.5 = 66.15N. Y component of gravity is mgcos30 which is 57.4N. The additional component of force...

Yes, OK - Fdg would be mgsinθ ; Fng would be mgcosθ ; and the frictional force due to Fng would be equivalent to the normal force since the box isn't moving...I am not positive about this last one but I am thinking since there is no motion the frictional force is that due to the box weight only.

Thank you - that makes perfect sense. I didn't see that. I am finding that my biggest obstacle seems to be that I have difficulty seeing everything that is there. I appreciate your help!

Homework Statement A 2.1 X 10^3 kg car starts from rest at the top of a driveway that is sloped at an angle of 20.0 degrees with the horizontal. An average friction force of 4.0 X 10^3 N impedes the car's motion so that the car's speed at the bottom of the driveway is 3.8 m/s. What is the...

OK - I am very new to this and am trying to explain it to others. I seem to be missing a key point here in the concept. I understand what you are saying but have no idea how to calculate this additional force. Please get me started.

OK - so am I misinterpreting the question?? Is there an additional force that must be applied to keep the box from slipping? I am afraid that I don't know what you are saying. I thought that adding the downward slope force in was a mistake but am really confused as to what to do...

Yes - the simplest approach would be to use the formula that equates the coefficient of static friction to static friction force divided by the normal force. Static Friction Force = 0.500 X mgcos30 = 28.7 N Another thing I thought made sense is to add the force due to the mass of the box...

Homework Statement The coefficient of static friction between the box (6.75 kg) and the ramp (30 degree incline) is 0.500. What is the magnitude of the minimum force, F, that must be applied to the box perpendicularly to the ramp to prevent the box from sliding? Homework Equations...