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3 years ago

11

At one particular moment, a 15.0 kg toboggan is moving over a horizontal surface of snow at 4.40 m/s. After 7.50 s have elapsed,

the toboggan stops. Use a momentum approach to find the magnitude of the average friction force (in N) acting on the toboggan while it was moving.

1 answer:

MArishka [77]3 years ago

7 0

Answer:

$f = 8.8\,N$

Explanation:

The moment equation of the toboggan is:

$m_{t}\cdot v_{o} - f\cdot \Delta t = 0$

The average friction force is:

$f=\frac{m_{t}\cdot v_{o}}{\Delta t}$

$f = \frac{(15.0\,kg)\cdot (4.40\,\frac{m}{s} )}{7.50\,s}$

$f = 8.8\,N$

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$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

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The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

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In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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