# Physics people I need your help

Ok this problem i will be shoing you is for extra credit. I cannot work with anyone from my class and i have already asked the teacher and it was a no go. So i figured i would you a loop hole and ask people outside my class (I am assuming you are’nt in my class )
Here it goes

A 25 kg child slides down a water slide starting from rest. The angle of the slide is 20 relative to the horizontal. The child slides friction free down the length of the incline which is 10 meters in length. The child then slides across a level water chute for 3 meters where the coefficient of friction between the child and the slide is .1 . At the end of the xhute is a 2 meter high drop off into the pool below. (A) How long is the child in the air one he leaves the chute? (B) How long does the entire ride take? © How far out horizontally from the end of the chute does the child hit the pool?

I asking you guys here because most of you seem very smart. I have tried making a free body diagram, but i had no idea where to start.

umm, yes we could solve this for you but what would be the point in that? you take classes to learn things. having people give you the answers is not really learning. besides, this is extra credit even.

This is a pretty basic physics problem so just work your mental muscles a bit.

EDIT:

well, in general, its good to split up the problem into its distinct parts (2 here). This also looks like something you would use Work/Energy for, then kinematics. that should be more than enough to get you started.

A 25 kg child slides down a water slide

This is all you need to know. It’s a trick question. Water slides have minimum weight restrictions, everyone knows that. Plus you must be taller than this sign.

23 seconds

guessing

What is with all of your posts and the big blank space?

Dude, you should see his spacebar.

Yes I have elephantitus in my spacebar Hmm, I kind of remember this kind of stuff. You would figure out his weight in newtons (25kg x 9.8m/ss), break that down into horizontal and vertical components. If I remember correctly the horizontal component would remain constant without friction. When it reaches the level area, the frictional force would be the coefficient of friction multiplied by the normal force. That would be subtracted from the horizontal component of force. That gives you a force that, when divided again by the weight, will give you the acceleration. And then . . . arrgghhh, pain in the brain.

A) 2 meters = 0 m/s x T + .5 x 9.8 m/ss x Tsquared. The answer is the square root of 2 divided by 4.9. I think.

B) Ah, I ain’t got time for this. Do what the rest of us do and learn it.

I know how to get the answers but I’m not going to bother working it out unless you do, then I’ll work it out and compare answers.

This is actually a three part problem.

Start by calculating the time it takes to get from the top of the slide to the beginning of the horizontal part (ignore the horizontal distance and calculate using the height only, I’ll explain later), you’ll also need the instantanious velocity here.

Calculate the time it takes to slide to the end of the horizontal part of the slide using the coefficient of friction (look it up in your book).

The velocity after the end of the slide is irrelevant for (a) since he will fall at a constant rate to the water, the time is given by: (9.8 m/s^2) / (2m). For part C use this amount of time multiplied by the velocity at the end of the slide for the horizontal distance travelled.

If you don’t believe me on the last part and the beginning about only calculating the vertical component for time consider this: A common physics experiment involves a stuffed animal and a projectile aimed at the stuffed animal, both hung at the same height. When the experiment is activated the stuffed animal falls and the projectile is shot at it from some distance away. The projectile always hits because they both fall at the same speed, the only thing that varies is the time it takes for the projectile to hit (though that is irrelevant for your problem).

The velocity at the end is very relevant, you’ll need it to get the horizontal distance travelled between leaving the slide and falling in the water for C.

Martin

The velocity at the end is very relevant, you’ll need it to get the horizontal distance travelled between leaving the slide and falling in the water for C.

Martin[/quote]

Ah crap, I just re-read that part. I thought it was asking for time.

My main problem is i dont have my freefall equations written down. Like the no time equation. I already know about breaking up the components of the forces not perpendicular to the way i orientated my axis. Also it gives the coeffiecient of friction in the problem. Teeth time is relevant because you need the time of the whole ride.

well, you could always rederive the formulas from calculus, hehe.

this problem is a lot simpler than I think you’re making it out to be. shbaz’s method works. An even easier way is to use the Work Energy theorem and you only need to use kinematics to calulate the time it takes to fall. In fact, with the WE theorem, you can roll the first two steps (frictionless slide and level chute) into one, assuming you know what it is/how to use it.

For the time of the entire ride you’ll need to split it into three parts and use kinematics and some trivial trig functions. also remember that the kinetic frictional force is constant, which means accel is constant, which means you can use kinematics on the chute.