An 1kg object is sliding down a 4m high incline which is inclined at 53 degrees. The initial speed is 2m/s. It then slides on a horizontal section 3m long which is at ground level and then slides up an incline plane that is inclined at 37 degrees. All surfaces have a kinetic coefficient of uc = 0.4. What distance will the object travel up the 37 degree incline before stopping?(adsbygoogle = window.adsbygoogle || []).push({});

Work Done:

Starting off Values:

Vi = 2m/s

Vf = ?

d = 3.19 (using trig)

m = 1kg

uc = 0.4

I drew a diagram and then I figured I would use the laws of conservation formula. I know their is kinetic and potential energy when the object is at a certain height. When the object is on the horizontal surface, their will only be kinetic and friction forces acting on it. So I thought i'd find the object's final speed on the 53degree incline (1/2mVf^2 = 1/2mVi^2 + mgh -fx). After I found Vf, I used that as the new initial speed for the horizontal surface. I then thought i'd find the final speed on the horizontal speed using the laws of conservation of energy (1/2mVf^2 = 1/2mVi^2 -fx). Finally, I took this value and used it as the new initial speed as the object is about to move up the 37 degree incline. I substituted it in the formula: mgh = 1/2mVi^2 - fx.

I keep getting an answer like 2.41m but the answer in the book is 1.95m. I'm not quite sure where I went wrong?

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# Homework Help: Energy Conservation

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