Dimensional Analysis: Solving E = (1/2) mv Equation

[Michele Nunes]

Homework Statement

Is the following equation dimensionally correct?

Homework Equations

E = (1/2) mv
where:
E = energy
m = mass
v = speed

The Attempt at a Solution

1. I understand that the 1/2 is irrelevant.
2. I broke everything down into length, time, and mass.
3. I got ML^2/T^2 = ML/T
4. My confusion is that if you square L and T on the left side, you still have length and time so I mean essentially, you have the same types of quantities on both sides of the equation in the same order, so I want to say it is correct, however the fact that L and T are squared on the left side but not on the right side bugs me and I'm doubtful.

 

[C. Lee]

Homework Statement

Is the following equation dimensionally correct?

By "the following equation" do you mean E=(1/2)mv^2 ?

 

[Michele Nunes]

By "the following equation" do you mean E=(1/2)mv^2 ?

I was referring to the equation under Relevant Equations (my apologies, should've clarified that), which is E=(1/2)mv

 

[C. Lee]

I was referring to the equation under Relevant Equations (my apologies, should've clarified that), which is E=(1/2)mv

You are missing ^2 at the end of the equation: it should be E = (1/2)mv^2.

 

[Michele Nunes]

You are missing ^2 at the end of the equation: it should be E = (1/2)mv^2.

Okay so it isn't dimensionally correct? The book just gives me an equation (it's not necessarily supposed to be right or wrong) and I'm just supposed to say whether the given equation is dimensionally correct. But since the equation itself is wrong then I'm going to assume that the correct version is dimensionally correct and this one is not.

 

[C. Lee]

Okay so it isn't dimensionally correct? The book just gives me an equation (it's not necessarily supposed to be right or wrong) and I'm just supposed to say whether the given equation is dimensionally correct. But since the equation itself is wrong then I'm going to assume that the correct version is dimensionally correct and this one is not.

Absolutely right. As you have pointed out from the beginning LHS has extra L/T that RHS does not have, therefore it cannot be dimensionally correct.