# Velocity at the bottom of an incline

1. Oct 10, 2004

### Tycho

How can I find the velocity of an object at the bottom of a frictionless incline?

m=108kg
Initial Velocity= 4.6m/s
Length of ramp=755m
Incline=downwards at 19.6 degrees

I keep telling myself i'm not unintelligent.... But this keeps replying otherwise :yuck:

2. Oct 10, 2004

### cepheid

Staff Emeritus
Use the given angle and the mass of the object to calculate the component of the object's weight acting parallel to the incline. With this net force acting on it (there is no friction), you know its acceleration from Newton's second law. Kinematics will tell you the final velocity of an object with the given initial velocity traversing this distance at constant acceleration.

3. Oct 10, 2004

### PrudensOptimus

This should be in the Highschool Help section, but here is how I would do it:

Given: mass, v0, dx, and angle theta, we want to find v.

There is a kenematics equation: v^2 = v0^2 + 2*a*dx

The only thing you do not know is the acceleration, which you can find from the given quantities. The only thing acting on your object is Gravity.

4. Oct 10, 2004

### Tycho

dx? is that the length of the incline?
and to find the acceleration, is that a=-g*SIN(angle)?

5. Oct 10, 2004

### PrudensOptimus

dx is the length of the line in this case.

Acceleration is defined to be a = Sum of all forces / m for constant mass objects.

your a = -g*SIN(angle) is misleading. It will not work in all cases.

6. Oct 10, 2004

### PrudensOptimus

Final velocity should be 70.6 m/s down the slope.

7. Oct 10, 2004

### aekanshchumber

Acceleration down the plane a = gsin(19.6)
distance, s = 755
initial velocity, v' = 4.6
substitute values in
V^2 = (V')^2 + 2as
v=71 m/s (aprox)

8. Oct 10, 2004

### PrudensOptimus

just in case you might ask why is the final velocity almost 15 times faster than the initial speed: The distance played a major factor in determining this such increase... as you go further, your speed increases.

9. Oct 11, 2004

### Tide

Why not just use energy conservation? The change in kinetic energy in this situation depends only on the net vertical displacement of the object. Remember Galileo! The rest is basic geometry.