# Calculating power with force as a function of time [closed]

While doing an AP question in Physics C today I answered the question differently from the professor but I'm not sure what part of my reasoning is incorrect.

A $$100\ \rm{kg}$$ block is being pulled along the x-axis with a position as a function of time of $$x(t)=0.5t^3+2t$$ Determine the power $$P(t)$$ being delivered to the box as a function of time.

The professor used $$P=Fv$$ to answer the question, but I wanted to attack the problem a different way to test my understanding of the physics concepts involved.

My method: $$F=m\ddot x=300 t$$ $$\text d x=(1.5t^2+2)\text d t$$ $$P=\frac{\text d}{\text d t}\int_0^xF\text dx'=\frac{\text d}{\text d t}\int_0^{0.5t^3+2t}(300t)(1.5t^2+2)\ \text d t$$

But working this out does not lead to the correct answer: $$P=450t^3+600t$$

Am I not understanding how $$P=\frac{\text d W}{\text d t}$$?

## closed as off-topic by ZeroTheHero, John Rennie, stafusa, Cosmas Zachos, M. EnnsNov 19 '18 at 1:05

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• The question is both about a specific physics concept and shows an attempt at the solution so why was this put on hold? – lepton10 Nov 19 '18 at 4:10

The issue comes down to how you are doing your change of variables in your integral. It turns out that you are changing the integrand correctly, but you are not changing the limits correctly.

The original work integral is $$W=\int_0^{x'}F(x)\text d x$$ So, the substitution you want is of the form $$t=g(x)$$ for some function $$g$$. Therefore, your lower limit becomes $$g(0)=0$$ and your upper limit becomes $$g(x')=t'$$. So you don't replace $$x$$ with $$x(t)$$, because that is not how you do the variable substitution. This is how you end up "double substituting" the position/velocity into the integral in the work you give as Gabriel pointed out in the comments.

In comparison to your professor's work, what you want to end up with and use is $$W=\int_0^{t'}F(t)v(t)\ \text d t$$ So that $$P=\frac{\text d W}{\text d t}= F(t)v(t)$$

In your work you end up with the right integrand $$Fv\text d t$$, but your upper limit just becomes $$t$$ because of this, not $$x(t)$$.

As a small aside: even though you are not looking for tips about the AP exam, just note that your professor's method is better for the exam due to how concise it is. However, it is also true that education is far more than the AP exam, so I hope that my answer has aided you in your educational journey. Curiosity in thinking about problems in different ways should never be stifled.

For more in depth detail of how the substitution works:

Let's first review the substitution process. We have an integral $$I=\int_{x_1}^{x_2}f(x)\ \text d x$$, and we define the substitution $$u=g(x)$$ so that $$\text d u=\frac{\text d g}{\text d x}\text d x$$. Therefore, the integral becomes:

$$I=\int_{g(x_1)}^{g(x_2)}\frac{f(g^{-1}(u))}{\left.\frac{\text d g}{\text d x}\right|_{x=g^{-1}(u)}}\ \text d u$$

Let's do the same thing in our problem. We have $$W=\int_0^{x'} F(x)\ \text d x$$

Let's make our substitution be $$t=g(x)$$, or $$x=g^{-1}(t)=h(t)$$. Then $$\text d t=\frac{\text d g}{\text d x}\text d x$$. But, if we use from calculus $$\frac{\text d}{\text d x}f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$$ what we actually get with our substitution is $$\text d t=\frac{\text d g}{\text d x}\text d x=\frac{\text d}{\text d x}h^{-1}(x)\cdot\text d x=\frac{1}{h'(h^{-1}(x))}\text d x=\frac{\text d x}{v(t)}$$ since $$h^{-1}(x)=t$$, and $$h'(t)=\frac{\text d h(t)}{\text d t}=\frac{\text d x}{\text d t}=v(t)$$.

So our work integral becomes: $$W=\int_{g(0)}^{g(x')}F(g^{-1}(t))v(t)\text d t=\int_{0}^{t'}F'(t)v(t)\text d t$$ where I have defined a new function $$F'=F\circ g^{-1}$$ that represents the force as a function of time (so in your case $$F'(t)=300t$$).

We have reached what we wanted. This work might seem convoluted (and I agree with that), but I have included this just in case you wanted to see step by step how you really were just doing a typical "u-substutution", but the substitution is of the form $$t=g(x)$$ rather than just $$x=x(t)$$ and $$\text d x=v\text d t$$.

• One comment: students only have an average of 15 minutes per problem on the physics C exam. This is only one part of a 6 part problem. – David White Nov 17 '18 at 6:07
• @DavidWhite, please stop talking about the AP exam. The OP has already said that this question is not intended to help with the AP exam. Nowhere in my answer do I say that this is the best way to do the problem, nor do I say this is how you should tackle problems like these on the AP exam. But if it makes you feel better I'll add a comment about the AP test. – Aaron Stevens Nov 17 '18 at 6:13