- #1

Eyris

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The mass is pushed up an incline with an angle of 1 degree with an initial velocity of 1 m/s, and it comes back down to its original position.

The questions to answer:

What is the total distance the object travels on the frictionless inclined plane?

How long will the object take to return to its starting position?

__Relevant equations:__

V = Vi + at

V^2 = V^2i + 2a(X-Xi)

__Knowns:__

Angle = 1 degree

Initial Velocity Up Plane = 1 m/s

Final Velocity at top of plane = 0 m/s

a = g*sin(angle)

__My attempt:__

My positive X axis is pointing up the incline, and my positive Y is perpendicular to the mass in the up direction.

**What is the total distance?**

d = (X-Xi)

The kinematic equation I will use to find the distance from bottom to top of incline:

V^2 = V^2i + 2 * a * d

a = -9.81 * sin(1degree) = -.171 m/s^2

V = 0 m/s (at the top of incline, velocity is 0)

Vi = 1 m/s

Plug in a, v, and vi into v^2 equation to solve for d:

0 = 1 + 2(-.171)*d

d = 2.92m (to the top)

2d = total distance

**What is the total time?**

total time = time to top + time to bottom

To the top

V = Vi + at

a = -.171 m/s^2

t = (V-Vi)/a = -1/-.171 = 5.85s (to the top)

To the bottom

Vi = 0 (at top of plane)

d = 2.92m

a = +.171 m/s^2

V^2 = 0 + 2(.171)(2.92) -- (Am I proceeding correctly?)

V = 1 m/s

*(would this actually equal -1; if so, is my previous V^2 equation incorrect?) V WHEN IT COMES BACK TO ITS ORIGINAL POSITION*

V = Vi + at

1 = 0 + (.171)(t)

t = 1/.171 = 5.85s (time to go down)

total time = 5.85 * 2 = 11.7s