# Block sliding down ramp

1. Jun 29, 2004

### akatsafa

A 10 kg block slides from rest down a 5m long ramp. If the coefficient of friction between the block and the ramp is 0.4, what is the final velocity of the block when it reaches the botton of the ramp?

I set this into two Fnet equations, x and y. I then solved for the acceleration in x which i got to be -3.92 m/s^2. However, I'm not getting the correct value for the final velocity. I'm getting 6.3 m/s, but it should be 3.9 m/s. Can you please tell me what I'm doing wrong?

2. Jun 29, 2004

### akatsafa

The angle at the bottom of the ramp is 30 degrees.

3. Jun 29, 2004

### Staff: Mentor

For one thing, your acceleration down the ramp is incorrect. Start by showing those equations for Fnet and how you solved them.

4. Jun 29, 2004

### akatsafa

I have fnetx=ma=-uN....I have fnety=0=N-W..I then tried solving for acceleration...that's how I got -3.92.

5. Jun 30, 2004

### Parth Dave

Make sure you set up your axes such that the x-axis is the same as surface of the ramp and the y-axis perpendicular to it. This way it is much easier, you only have to break Fg into it's components. I'm not quite sure what you've set up for the net force in the x and y direction, however i notice that in the net force for the x direction you do not have gravity included. Here is what they should be however:
Fnetx = ma = Fgx - Fk(uFn) (force of gravity in the x direction - force of kinetic friction)
Fnety = ma = Fn - fgy (Normal force - Force of gravity in the y direction)

Also, im not sure why you are given mass.. as you dont need it.

6. Jun 30, 2004

### KnowledgeIsPower

Angle of 30 degrees, U = 0.4, mass = 10kg, distance = 5m.

Resolving parallel force: 10GSin30 = 49N
Resolving opposite to plane, force: R - 10Gcos30 = 0, R = 84.87N
F = UR, F = .4x84.87 = 34N

49 - 34 = 10A
A = 1.5m/s/s.

u = 0, v = ?, s = 5, a = 1.5
v^2 = u^2 + 2as
?^2 = 2x5x1.5
V = 3.873 m/s.

Last edited: Jun 30, 2004
7. Jun 30, 2004

### Staff: Mentor

Assuming "x" means "parallel to the ramp" and "y" means "normal to the ramp", the forces in the x direction are the x-component of the weight (mg sin(30)) and the frictional force. So:
$$\sum F_x = -mg sin(30) + \mu N = ma$$
And the forces in the y direction are the y-component of the weight (mg cos(30)) and the normal force. So:
$$\sum F_y = -mg cos(30) + N = 0$$
Solve for N, then for a. As has been pointed out, the mass drops out and is not needed.