a = gsin(\theta) - \mu kgcos(\theta) is correct.
Which parts do you need help with understanding? Or do you not understand any of it?
First sums up all the forces acting on the child, then finds that the net force to allow the child to move is in the direction parallel down the slide.
He...
I'm not entirely sure why that is, but it could have something to do with order of operations (don't quote me on that). Using your formula, I would take a reciprocal of it first: R + r = \frac{2GM_e}{-v^2} + R
From this we can see that the R terms cancel and we're left with r that we want to...
Hmm you will want to get a second opinion on this then, I interpreted the question as solving for the acceleration in the x plane (possibly because of the axes they provided).
Bingo!
We also know that Fg = mg (right?)
Therefore, we can say that F_{parallel} = mgsin(\theta)
We also know that F = ma. Therefore, mgsin(\theta) = ma and we are solving for a (acceleration). Can you solve this?
Not at all, your picture is correct. Here is a modified version of it:
http://img265.imageshack.us/img265/2159/webassignplanesa0ya6.jpg
pardon the poor drawing skills.. An important concept (but sometimes hard to get a grip of initially) to realize is that just as the three sides of the...
We're not at the answer yet. This is just the normal reaction force. It is infact F_gcos(\theta) . This is because the if Fg is the hypotenuse, Fn is the adjacent angle to theta. Using SOHCAHTOA we can derive F_gcos(\theta) .
Now, we have the normal reaction force and the gravitational force...
Okay, we have to understand that we're not solving for numbers yet. We are just solving algebraically, using letters to represent the forces.
I can see where you're a little confused, Fg = mg, but we don't know m. (g of course = 9.8ms-2)
Try to solve for Fn in terms of Fg using trigonometry...
No, Fn is going to be the component of gravity that is perpendicular to the plane. We have the hypotenuse (Fgravity) and we have an angle (25 degrees). Can you use trigonometry to find Fn in terms of Fgravity? Let me know if you aren't understanding this too.
Okay, first we'll write this problem out algebraically.
You need to use your basic knowledge to find out where the forces are in this question.
We can assume that there is going to be a force on the mass due to the acceleration due to gravity. Fgravity = mg.
There is going to also be a normal...
Ahh, but you don't need to know what the mass is to solve it! Take the advice from my first post; sum up the forces in each direction and tell me what you get.
You do need to use right angled trig, and you do have enough information.
I would start by defining all of the forces acting on the block (a free body diagram can help).
Then sum up the forces in each direction, divide by mass and you have your answer.
Sorry, but this is really hard to follow. I find it much easier to write out the whole problem algebraically before subbing in any numbers, and also using consistant significant figures.
If you could do this first it would help a lot.
So, we know that energy is going to be conserved. First I'd write out a balanced equation using all the types of energy we have in this question:
\frac{1}{2}mv_1^2 + F_1\Delta d_1 = \frac{1}{2}mv_2^2 + F_2\Delta d_2
because there is no initial friction, we can get rid of the + F\Delta d...
It would help to show what you've tried so far,
We know that momentum is conserved (mass * velocity) within a system. Initially I assume that none of the astronauts are moving. What does this tell you about the motion of the astronauts after the first throw?
Don't worry about it, we're here to help you haha! Don't hesitate to post again if you still can't solve it. Sorry that it took so long to understand your question.. I'm terrible at questions without a diagram infront of me..