Can I get some Physics help?

Post » Thu Dec 08, 2011 12:40 am

Believe me, if I could do this stuff, I would. Thanks in advance to all who help answer, attempt to help, or nudge me in the right direction. Now, I don't want you guys to think that I want you to do my homework, I don't need the answers I just need a nudge in the right direction, like even the formula or something that I'm supposed to use would be good. Okay, so here they are:

1. A 64-kg person on skis is going down a hill sloped at 30° from the horizontal. The coefficient of friction between the skis and the snow is 0.12. What would be the magnitude of the acceleration?

2.
A. A 60-kg person on skis starts from rest down a hill sloped at 36° from the horizontal. The coefficient of friction between the skis and the snow is 0.20. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero. What is the new coefficient of friction?
B. How fast would the skier now be going after skiing for another 5.0 s?

3. A 240-N box is placed on an inclined plane that makes a 40.0° angle with the horizontal. Find the component of the weight force parallel to the plane's surface.
User avatar
Sunnii Bebiieh
 
Posts: 3454
Joined: Wed Apr 11, 2007 7:57 pm

Post » Thu Dec 08, 2011 1:23 am

Haven't done any physics problems in about two years now, so I could be wrong but!

Believe me, if I could do this stuff, I would. Thanks in advance to all who help answer, attempt to help, or nudge me in the right direction. Now, I don't want you guys to think that I want you to do my homework, I don't need the answers I just need a nudge in the right direction, like even the formula or something that I'm supposed to use would be good. Okay, so here they are:

1. A 64-kg person on skis is going down a hill sloped at 30° from the horizontal. The coefficient of friction between the skis and the snow is 0.12. What would be the magnitude of the acceleration?

F = ma, Ff = (coefficient)(normal force)
where F = sum of forces, Ff = force of friction
Remember that the normal force is perpendicular to the surface (that's where the 30 degrees comes in play)

Forces in play are: Gravitational force (mg), frictional force, and normal force

See if that helps.
2.
A. A 60-kg person on skis starts from rest down a hill sloped at 36° from the horizontal. The coefficient of friction between the skis and the snow is 0.20. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero. What is the new coefficient of friction?
B. How fast would the skier now be going after skiing for another 5.0 s?

Uhhh I honestly don't remember enough. I'll look at it in a few minutes.

3. A 240-N box is placed on an inclined plane that makes a 40.0° angle with the horizontal. Find the component of the weight force parallel to the plane's surface.

I'd have to draw this out to explain it. Sorry. But remember that that particular force is composed of two parts, the x-component and the y-component (where negative y is in the same direction as gravity and positive x is perpendicular and to the right of y). The rest is trigonometry and knowing the other forces.

Hopefully that last part made sense. I find it always helps to draw out concrete problems like these.
User avatar
GEo LIme
 
Posts: 3304
Joined: Wed Oct 03, 2007 7:18 pm

Post » Thu Dec 08, 2011 9:16 am

Thanks for the help! I'll try those things very soon. Hopefully I'll get them right this time.

Alright I got the first one.

Awwwwwwww right, got the second one. Now to try my hand at the third one.

Nevermind, got it them all now then.

Thanks for your help trira!
User avatar
Chantel Hopkin
 
Posts: 3533
Joined: Sun Dec 03, 2006 9:41 am


Return to Othor Games