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

AI Thread Summary
The equation E = (1/2) mv is incorrectly presented, as it should be E = (1/2) mv^2 to be dimensionally correct. The left-hand side (LHS) has dimensions of ML^2/T^2, while the right-hand side (RHS) has dimensions of ML/T, leading to a mismatch. The confusion arises from the presence of squared terms on the LHS that are absent on the RHS. Consequently, the original equation is not dimensionally correct. The correct version, E = (1/2) mv^2, maintains dimensional consistency.
Michele Nunes
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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.
 
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Michele Nunes said:

Homework Statement


Is the following equation dimensionally correct?

By "the following equation" do you mean E=(1/2)mv^2 ?
 
C. Lee said:
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
 
Michele Nunes said:
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.
 
C. Lee said:
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.
 
Michele Nunes said:
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.
 
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