I have no idea how to solve this physics problem :(

Start by breaking out the forces acting on Willie. You have gravity pulling him down. The ramp holding him up and friction retarding his motion.

1. Break out the gravity force into that force normal to the ramp surface and parallel to the ramp surface. That is what the 55 degree number is for. The normal force is total gravity times the cosine of 55 degrees. The parallel force is gravity times the sine of 55 degrees. The ramp will completely counter the normal force, leaving only the parallel force to propel Willie.

2. Calculate the friction resistive force. Here you need the normal force calculated above again since friction is based on the normal force between the skier and the hill. The resistive force pushes directly against the parallel force from above, so now you know the total force propelling Willie down in the direction of the hill.

3. Find Willie's speed at the end of the down ramp. This is your standard distance = 1/2 a t^2 and velocity = at type problem. The picture and the wording don't agree here. When you say Willie travels 45m, I would take that to mean 45m on the ramp, but the drawing shows 45m horizontally which means you have to calcualte the ramp distance.

4. Convert that velocity into vertical and horizontal components using the 25 degree ramp, calculate the amount of time it takes Willie to hit the ground in the vertical direction then find out how far he travelled horizontally.

OK, these are pretty broad hints I know, but they should get you started. If you try to work it through with these hints and post back what you get.

thanx alot
Hey engineer I can't tell you how great it felt to get a reply
I posted this problem in a physics forum but I got 0 replies.
I really appreciate that you took your time to read this and help.

It turns out that u r right about #4 45m is the length of the ramp I guess I labeled it wrong

Anyway first I started with a formula I used to work with W=MG SO I PLUGGED IN THE VARIABLES AND GOT W=65(9.8) WHIC EQUALS W=637

GRAVITY OR G=9.8 RIGHT??? or should it be -9.8?

I used the formula you suggested to get the normal force and the parellel force

Fn=9.8cos55 Fn=5.62N
Fp=9.8sin55 Fp=8.028N

I also used a formula I know Ff=-umg Ff=-.01(637) Ff=-6.37N


I stopped here cause I assumed that I did something wrong because as I mentioned earlier I'm not sure about the gravity.
Also, the formulas you mentioneed something earlier about "the ramp completely counter the normal force , leaving only the parallel force to propel willie" I didn't really get that.

Sorry to bother you again..but if you could please post back and tell me if I'm on the right track.

thanx a billion

Interesting. I actually had to learn a new section of physics to attemp this. I'm high school, this is most definetily university. Heres what I tryed. Remember this is merely an attempt.


Page 1


Page 2


As you can see in the first one I calculated the Fk because the body is moving therfore must have kinetic friction.

In the second one I calculated the acceleration, resultant force and velocity. I'm not sure where to go from here because of that Upward arced ramp. Well I tryed, might be a good attemp, might be crap, lol.

I would imagine his speed at exit just before he hits the 25 degree up lip was using v^2 - u^2 = 2as, and allowing for coefficient of friction and slope of hill...

v^2 - 0 = 2 * 45 * (9.81 * sin 55 * (1 - 0.01))

but please check how I used coefficient of friction as a retardent in that formulea, I believe it is multiplicative...

Which gives you a total speed of about v = 26.76 m/s.

So his vertical speed will be 26.76 * sin 25 and his horizontal speed 26.76 * cos 25 right at the exit. So he will travel up for 26.76 * sin 25 /9.81 = 1.15 seconds, reaching a height of 95 + ( 26.76 * sin 25/ 2 * 9.81) = 101.52 metres. So he falls this distance in s = ut + 1/2 a t^2 -> (as u = 0) t = sqrt(101.52 *2 / 9.81) = 4.55 seconds

So from takeoff he travels at a constant hozironatal velocity for 4.55 + 1.15 seconds so he travels horizontally (dh) 26.76 * cos 25 * (1.15 + 4.55) = 138.24 metres

Aaah, I see. Wow, this is a great question. I got the speed as 26.78 so I must have done the first part correctly. Thanks for the 2nd part, I spent the better part of the morning on this question with no result for the distance travelled.

Good start
OK, you got the gravity part. It doesn't matter if you use 9.8 or -9.8 as long as you are consistent. So the force of gravity that is propelling Willie down the hill is (8.03 m/sec^2)(65kg). The normal force on the ramp is (5.62 m/sec^2)(65kg). The resistive force due to friction is mu*Normal force. Mu=.01, so the force resisting motion is .01*(5.62)(65). For right now, let's just call that (.0562)(65). This means the total force driving Willie down the ramp is (8.03)(65) - (.0562)(65) = (7.97 m/sec^2)(65kg). The reason I haven't multiplied all of this out is that the important number is the acceleration, not the total force. Willie is seeing 7.97 m/sec^2 acceleration.

The key equation for the next part is:

X = Xo + VoT + 1/2 A T^2

which is the formula for position under constant acceleration. Willie is going 45 m, so X is 45 and Xo = 0. His initial velocity (Vo) is also zero so you are left with 45 = A T^2. Now we can calculate T to be sqrt(2x45/7.97) or 2.38 sec. If he accelerates at 7.97 m/sec^2 for 3.36 sec, he will be going 26.78 m/s. That is his velocity at the bottom of the ramp.

That velocity is going to be redirected by the ramp onto a 25deg upward angle. The problem is now a standard bullet fired from a gun problem. You need to break the velocity into an upward component, V*sin(25deg) and a horizontal component V*cos(25deg). Use the equation above to figure out how long it takes to Willie to hit the ground. Vo = the upward velocity you just calculated, A is the force of gravity and here you must use the negative sign. (X - Xo) = -95 meters since he ends up 95 meters below where he starts. Solve the quadratic for time. The distant he travels in the horizontal direction can now be calculated using Vh*T.

Post back if that doesn't get you there.

I was just looking at the other replies and GDay has pretty much nailed this for you, including the use of a more efficient formula, except for the application of friction. Friction is dependent on the normal force vector, not the vector in the direction of motion. Instead of substracting Mu from one, you have to substract Mu/tan(Theta). You can show that by looking at what happens as the angle approaches 90 degrees. At that point, there is no friction at all (Mu/tan(theta) -> 0).

wait how did you get 101.52 m as the height. Cause accourding to those calculations its 95 + (55.47) = 150.47

So thats basically a more complex version of the formula s = ut + 1/2at^2?

It looks like the same formula used twice, once to the peak and then to the bottom. You could do it in one step like you said.

Red888

He leaves the ramp with total speed around 26.78 m/s - of which vertical speed will be that times sin 25 = 11.32 m/s. At this point he starts accelerating due to gravity - so his vertical speed will be zero after 11.32 / 9.81 seconds = 1.15 seconds.

Using the very handy v^2 - u^0 = 2as formulea (with v = 11.32, u = 0 , a = 9.81 and solving for s shows he climbs vertically from 95 another 6.25 metres to 101.25 metres.

A reasonableness check - if your vertical velocity is only 11m/s and you decelerate at almost 10 m/sec/sec - no way will you climb 50m - about 5 is all you'll manage.

thanx u guys
I'm just starting to work on this but I don't really get these formulas??
I have never seen them before

Hmmm, ok thanks g_day. Makes you think.

Check it out rowntrees. There is quite abit of end level high school physics here but the co effecients of friction was not in my high school syllabus. I don't know about yours. I searched for it on google and found some tutorials. I learnt some of the basics there. It looks complicated but it's actually quite easy to pick it up.

Okay some formulea come from year 11 physics and some typically from year 11 engineering (dynamics)

Physics

s = distance
u = intial velocity
v = final velocity
a = acceleration
F = force
m = mass
h = heigth
p = momentum
KE = kinetic energy
PE = potential energy
W = work

* * * Standard formulea * * *

F = ma

p = mv

s = ut + 1/2at^2
v^2 - u^2 = 2as

PE = mgh (for an object about to drop a distance h)
KE = 1/2mv^2 (kinetic energy of a body in motion)

W = F*s (force * distance = work)

Dynamics

Body sliding down a hill recieves the force of sin x of the slope * gravity

Body sliding (not rolling) down a hill with friction (of the hill - not the air) recieves a force of sin x * gravity - coeffecient of (sliding) friction

There is also a lesser coefficient of rolling friction.

When Potential energy gets converted into kinetic energy (motion) (or vice versa) then energy is always conserved unless it is stated some leaks out as heat or noise etc.

Momentum and kinetic energy are always both independently conserved in any collision!

So drop a 1 kg ball 10 metres and its potential energy f = mgh = 1 * 9.81 * 10 = 98.1 Netwons gets converted into equivalent kinetic energy = 1/2 m v^2 = 1/2 * 1 * v^2, so v = sqrt (2 * 98.1) = 14 m/sec

Check v^2 - u^2 = 2as v^ - 0^2 = 2 * 9.81 * 10 so v = 14

The v^2 - u^2 = 2as is very handy if you have a starting speed and ending speed, don't know time but know either acceleration or distance.

e.g. a trucking weigthing 10,000kgs travelling 100 metres/sec has to stop in 300 metres - how much force must be applied?

Well f = ma, m = 10,000kg so find a

v = 100, u = 0, s = 300 so using v^2 - u^2 = 2as,

a = (100^2 - 0^2)/(2 * 300) = 16.7 m/sec/sec

So force required is 10,000 * 16.7 = 167 Kilo Newtons is required to stop the truck.

Hope this helps!

Another very useful formula is that of impulse.
FΔt = mΔv

So F = m( v - u ) / t

This was the actual forumla used by Isaac Newton in his manuscripts. You know fo cource that (v - u) / t = a That is how he derived F = ma. But that base forumla
FΔt = mΔv is very useful in physics.

il7imdilla- I think I got it
At last I worked on this all night > here is the final work It's five scans altogether I hope I have it right
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phew

Yup, that pretty much does it.

Gratz.

Did I use all the right formulas?
In the first part is my diagram right I mean with the forces

I just wanted to check isn't Vv=costheta
someone mentioned that it was sin theta

thanx alot for everything