# Speed at an inclined plane

1. Feb 14, 2008

### zcabral

1. The problem statement, all variables and given/known data
A block slides down a frictionless plane having an inclination of = 12.2° (Fig P5.22). The block starts from rest at the top and the length of the incline is 2.85 m.

2. Relevant equations
F=ma
xf=xi+vit+1/2at^2

3. The attempt at a solution
i tried to solve this by using the kinematic equation above. i know the acceleration is 2.07 m/s^2 so i plugged it in to find t in order to find speed but it didnt work. im totally stuck. i know if i find t i could find the answer by absolute value v=dx/dt

2. Feb 14, 2008

### dabig2

If you're searching for the velocity of the block at the very bottom of the ramp, you actually utilize Energy conservation instead of kinematics.

PE = mgh
KE = 1/2*mv^2

The block is at rest on top of the incline plane, which means no KE and only PE. That means it starts out with a total E of m*9.8*(2.85*sin 12.2)

At the bottom of the ramp, all that PE has been converted into KE.

KE = 1/2(m)(v^2)

So since E is conserved because there is no friction on the ramp nor air drag, set PE = KE and solve for V. Note that your masses will cancel out.

3. Feb 14, 2008

### zcabral

the mass is not given

4. Feb 14, 2008

### Littlepig

no problem, if the mass is the same in the begin and in the end, they cut in the equation... state the equation on a paper and you will see

5. Feb 14, 2008

### Oerg

although its much easier to do by consrevation of energy, but you should also reflect on why kinematics did not work for you. Are you still weak in kineatics? You may have escaped from using kinematics from this question but not all questions can be solved by conservation of energy,

The correct equation to use would be v^2=u^2+2as. if you used the equation in your OP, then you should have solved quadratically for t and substituted t into (v-u)=at.

6. Feb 14, 2008

### physixguru

this should be enough....

velocity at the bottom of an inclined plane is independent of angle of inclination...

infact it is given by...

v=[(2)gh]^1/2

Last edited: Feb 14, 2008
7. Feb 14, 2008

### Oerg

I think it would be way better to derive this result rather than to mug this equation and plug the numbers into it

8. Feb 15, 2008

### physixguru

well you are right but is it a too difficult equation my frnd????