Question Video: Finding a Formula for the Ratio of Two Quantities | Nagwa Question Video: Finding a Formula for the Ratio of Two Quantities | Nagwa

Question Video: Finding a Formula for the Ratio of Two Quantities Physics

Kinetic energy is given by the formula 𝐸 = 1/2 π‘šπ‘£Β². The weight, π‘Š, of an object is mg. Which of the following is a correct formula that has a subject equal to the ratio of the kinetic energy of an object to its weight? [A] 𝐸/π‘Š = 2𝑣²/𝑔 [B] 𝐸/π‘Š = 𝑔𝑣²/2 [C] 𝐸/π‘Š = 2𝑔/𝑣² [D] 𝐸/π‘Š = 𝑣²/2𝑔 [E] 𝐸/π‘Š = 𝑔/2𝑣²

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Video Transcript

Kinetic energy is given by the formula 𝐸 equals a half π‘šπ‘£ squared. The weight, π‘Š, of an object is π‘šπ‘”. Which of the following is a correct formula that has a subject equal to the ratio of the kinetic energy of an object to its weight? A) 𝐸 over π‘Š equals two 𝑣 squared over 𝑔. B) 𝐸 over π‘Š equals 𝑔𝑣 squared over two. C) 𝐸 over π‘Š equals two 𝑔 over 𝑣 squared. D) 𝐸 over π‘Š equals 𝑣 squared over two 𝑔. Or E) 𝐸 over π‘Š equals 𝑔 over two 𝑣 squared.

Since this question is asking us to find the ratio between two quantities, let’s start by recalling that the ratio of two quantities is the first quantity divided by the second quantity. So, for example, the ratio of 𝐴 to 𝐡 is just 𝐴 divided by 𝐡. Since this question asks us to find the ratio of the kinetic energy of an object to its weight, that means we’re looking for kinetic energy divided by weight. The question tells us that kinetic energy is represented by the symbol 𝐸. And weight is represented by the symbol π‘Š. So kinetic energy divided by weight is 𝐸 over π‘Š. And sure enough, we can see that all of our available answers start with 𝐸 over π‘Š.

So, we’re looking to find a formula that has a subject equal to 𝐸 over π‘Š. That’s the ratio of kinetic energy to weight. And the other side of the formula should look like one of these expressions. The question tells us that kinetic energy, 𝐸, can be expressed as a half π‘šπ‘£ squared. And it also tells us that weight, π‘Š, can be expressed π‘šπ‘”. That means we can rewrite a ratio 𝐸 divided by π‘Š by replacing 𝐸 with a half π‘šπ‘£ squared and replacing π‘Š with π‘šπ‘”. Now, what we have here is actually what the question is asking for. It’s a correct formula. And the subject is equal to the ratio of the kinetic energy of an object to its weight. However, we have a bit more work to do before it matches one of the available answers.

If we look at our expression, there are a couple of signs that it hasn’t been fully simplified. Firstly, on the right-hand side of the equation, we have a fraction with a fraction inside it. Now, this can look a bit confusing. And it’s always possible to rewrite fractions within fractions as a single fraction. So whenever you see a fraction inside another fraction, it’s a good sign that the expression can be simplified further. The second thing that shows us that we haven’t fully simplified our expression is that we can see a factor of π‘š on the numerator and on the denominator. And if we look at our possible answers again, we’ll see that none of them actually have an π‘š in them at all.

Let’s deal with these π‘šβ€™s first. And then, we can take a look at the fraction inside a fraction issue. Because we have a factor of π‘š in the numerator and the denominator of this fraction, they actually cancel each other out. To show this, recall that you can rewrite a fraction by multiplying or dividing the numerator and the denominator by the same thing. In this case, we’ll divide them both by π‘š. So the numerator of this fraction is a half π‘šπ‘£ squared divided by π‘š, which just leaves us with a half 𝑣 squared. And the denominator of the fraction is π‘šπ‘” divided by π‘š, which just leaves us with 𝑔. So we’ve now simplified our expression so that it no longer contains an π‘š. When simplifying equations in this way, it’s common to just draw a line through any factors that appear on the numerator and the denominator, to show how the quantities cancel out.

Okay, so we’ve simplified our expression. But we still need to write the right-hand side as a single fraction so that it matches one of the possible answers. To do this, let’s use a slightly simpler example. Let’s say we have the expression π‘Ž times 𝑏 divided by 𝑐. There are a few different ways we can write this. For example, we can remove π‘Ž from the numerator of the fraction and write it like this. So we can take any factor from the numerator, in this case π‘Ž or 𝑏, and write it as a factor of the entire fraction. Similarly, we could take 𝑏 out of the numerator of the fraction and write the expression as 𝑏 times π‘Ž over 𝑐. We can apply this method to the right-hand side of our expression by taking out the factor of a half from the numerator, giving us a half times 𝑣 squared over 𝑔.

All we have to do now is multiply these two fractions together. And we’ll get a single fraction as a result. Recall that when multiplying fractions together, we can just multiply the numerators and the denominators separately. So in our numerator, we have one times 𝑣 squared. And in the denominator, we have two times 𝑔, which we can write as 𝑣 squared over two 𝑔. This is the simplest form that the right-hand side of our formula can be written in. And it now matches one of the available answers.

The formula that has a subject equal to the ratio of the kinetic energy of an object to its weight is 𝐸 over π‘Š equals 𝑣 squared over two 𝑔.

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