- #1

Venturi365

- 12

- 3

- Homework Statement
- An Atwood's machine uses two masses ##m_{1}## and ##m_{2}##. Starting from rest, the velocity of the two masses is ##v=4\,\mathrm{\frac{m}{s}}## after ##3\,\mathrm{s}##. In that instant, the kinetic energy of the system is ##E_{k}=80\,\mathrm{J}## and each one of the masses has moved ##6\,\mathrm{m}##. Calculate the values of ##m_{1}## and ##m_{2}##

- Relevant Equations
- ##P=\frac{E_{k}}{t}##

##P=F\cdot v##

##\sum F=ma##

##E_{k}=\frac{1}{2}mv^{2}##

I've tried to solve this exercise but I haven't used one of the properties of the system (the displacement of the masses) so I don't know if I'm wrong about my procedure.

First of all, we (obviously) know that

$$

P=P

$$

And since we can express the power of a force in two different ways, we can say that:

$$

\frac{E_{k}}{t}=F\cdot v \quad \text{or} \quad \frac{mv^{2}}{2t}=F\cdot v

$$

Since the energy of the system is known and the mass of the system is ##(m_{1}+m_{2})## then:

$$

\begin{split}

\frac{E_{k}}{t}&=\frac{mv^2}{2t}\\

\frac{80}{3}&=\frac{(m_{1}+m_{2})4^{2}}{2\cdot 3}\\

(m_{1}+m_{2})&=\frac{80\cdot 2\cdot 3}{16\cdot 3}\\

m_{1}+m_{2}&=10\,\mathrm{kg}

\end{split}

$$

Applying Newton's Second law we can express the forces on each mass like this:

$$

\begin{split}

w_{1}-T&=m_{1}a\\

T-w_{2}&=m_{2}a

\end{split}

$$

Adding up both equations we get:

$$

\begin{split}

m_{1}g-m_{2}g&=m_{1}a+m_{2}a\\

g(m_{1}-m_{2})&=a(m_{1}+m_{2}\\

(m_{1}+m_{2})a&=g(m_{1}-m_{2})\\

(m_{1}-m_{2}&=\frac{a(m_{1}+m_{2})}{g}\\

m_{1}-m_{2}&=\frac{\frac{4}{3}\cdot10}{9.81}\\

m_{1}-m_{2}&\approx =1.36\,\mathrm{kg}

\end{split}

$$

Now we have a simple system of equations:

$$

\begin{cases}

m_{1}+m_{2}=10\\

m_{1}-m_{2}=1.36

\end{cases}

$$

Solving we get:

$$

\begin{array}{ll}

m_{1}=1.36+m_{2} & m_{1}+4.32=10\\

1.36+m_{2}+m_{2}=10 & m_{1}=10-4.32\\

m_{2}=\frac{10-1.36}{2} & m_{1}=5.68\,\mathrm{kg}\\

m_{2}=4.32\,\mathrm{kg}

\end{array}

$$

So, am I right or did I do something wrong? The only thing that I really doubt about my procedure is the fact that I assume that the net force is ##F=a(m_{1}+m_{2})## but still I'm quite insecure.

First of all, we (obviously) know that

$$

P=P

$$

And since we can express the power of a force in two different ways, we can say that:

$$

\frac{E_{k}}{t}=F\cdot v \quad \text{or} \quad \frac{mv^{2}}{2t}=F\cdot v

$$

Since the energy of the system is known and the mass of the system is ##(m_{1}+m_{2})## then:

$$

\begin{split}

\frac{E_{k}}{t}&=\frac{mv^2}{2t}\\

\frac{80}{3}&=\frac{(m_{1}+m_{2})4^{2}}{2\cdot 3}\\

(m_{1}+m_{2})&=\frac{80\cdot 2\cdot 3}{16\cdot 3}\\

m_{1}+m_{2}&=10\,\mathrm{kg}

\end{split}

$$

Applying Newton's Second law we can express the forces on each mass like this:

$$

\begin{split}

w_{1}-T&=m_{1}a\\

T-w_{2}&=m_{2}a

\end{split}

$$

Adding up both equations we get:

$$

\begin{split}

m_{1}g-m_{2}g&=m_{1}a+m_{2}a\\

g(m_{1}-m_{2})&=a(m_{1}+m_{2}\\

(m_{1}+m_{2})a&=g(m_{1}-m_{2})\\

(m_{1}-m_{2}&=\frac{a(m_{1}+m_{2})}{g}\\

m_{1}-m_{2}&=\frac{\frac{4}{3}\cdot10}{9.81}\\

m_{1}-m_{2}&\approx =1.36\,\mathrm{kg}

\end{split}

$$

Now we have a simple system of equations:

$$

\begin{cases}

m_{1}+m_{2}=10\\

m_{1}-m_{2}=1.36

\end{cases}

$$

Solving we get:

$$

\begin{array}{ll}

m_{1}=1.36+m_{2} & m_{1}+4.32=10\\

1.36+m_{2}+m_{2}=10 & m_{1}=10-4.32\\

m_{2}=\frac{10-1.36}{2} & m_{1}=5.68\,\mathrm{kg}\\

m_{2}=4.32\,\mathrm{kg}

\end{array}

$$

So, am I right or did I do something wrong? The only thing that I really doubt about my procedure is the fact that I assume that the net force is ##F=a(m_{1}+m_{2})## but still I'm quite insecure.