Newton's 2nd Law problem (I think )

[frankfjf]

Alright, here's the problem:

A block is projected up a frictionless inclined plane with initial speed v0 = 7.917811929967 m/s. The angle of incline is = 36.8°. (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

I'm not sure how to solve frictionless inclined plane problems, but I gave it a shot and here's what I got:

Vi = 7.9
Theta = 36.8
g = 9.8

The above are of course the known variables, and it seems I'm attempting to solve for vertical distance, time, and speed on the way back.

Here are the variables I think I've solved for:

I wasn't able to solve for Normal Force since I'm unsure how that works.

m = ?
t(time) = .81s

Vf = 0 (I assume that since they want to know how far up the block gets, it won't make it all the way up and even if it does it'll come to a stop, so I'm assuming the final velocity is zero.)

However, when I plug in what I know and what I think I know into the formula:

d(distance/displacement) = Vi(t) + (1/2a)(t^2)

I do not get the right answer.

To solve for time I attempted to use the formula:

Vf = Vi + at

But I'm not sure if I'm following proper proceedure.

If not, are the formulas involved at least correct?

Other than that, do I need the formulas for projectile motion or no?

Thanks in advance.

 

[phucnv87]

For the a) problem, using the law of conservation of energy
For b), let write out the Newton's second law and find the acceleration of the object. Then the time is $$\Delta t=\frac{v_0}{a}$$
For c), doing as the a).[/color]

 

[frankfjf]

I don't understand what you mean..

 

[phucnv87]

For a), the final gravitational energy is

$$W=mgl\sin\theta$$

The initial kinetic energy is

$$K=\frac{1}{2}mv^2$$

The friction force is

$$F=\mu mg\cos\theta$$

The law of conservation of energy

$$K=Fl+W$$

And continue...[/color]