# Physics I

an object weighing 3.0 kg has an initial velocity of 3.33m/s up a 37 degree incline. how far will it go before it stops?

you don't have to do the work, I just don't know what equations to use to solve it.

What's conserved in this situation?

Nitpick:

Kilograms isn't a measure of weight.

Ok, so I'm new at this, so make sure to double check, but here are some ideas:

Method 1:

Conservation of Energy (Initial Total Energy = Final Total Energy)
Total energy is Kinetic Energy (Ek = .5mv^2) and Potential Energy (Ep = mgh)

Initial Energy = Final Energy
Ek + Ep = Ek' + Ep'

Initial Ep is 0 (since initial height, h, is 0 ) ; and final Ek is 0 (since final velocity, v', is 0)
Mass cancels (and is not needed unless we think about friction).
This gives us the height, h, which is the vertical distance - not your final answer. Draw a picture and use sin37 to find how far the object travels along the slope.

Method 2:
v'^2 = v^2 + 2ad (final velocity, v', is 0)
Mass is also not needed here! (Check the question to make sure that friction can be ignored.) Acceleration is not simply -9.8m/s^2. I imagined the slope rotated so that it is horizontal. Now find the acceleration provided by horizontal component of the force of gravity. I could also imagine the slope rotated vertically, and then find the acceleration provided by the component of the force of gravity pulling it down. The component of the force of gravity perpendicular to the slope (pulling the object into the slope) is countered by the normal force and can be ignored because we are not worried about friction.

My final answer was .94 m . Try doing it both ways and make sure the answers agree. Let me know if you see any mistakes I may have made... I am just learning this myself.

Just did it myself and my trusty calculator gave an answer of .94.When I then checked your answer I knew I got it right.