To find the sled's initial speed, we can use the work-energy principle. When the sled encounters the rough patch, friction is the only force acting to slow it down, so the work done by friction will be equal to the initial kinetic energy of the sled.

The work done by friction ($W$) can be expressed as: $W = F_{\text{friction}} \times d$ where:

• $F_{\text{friction}} = \mu \cdot m \cdot g$ (the frictional force),
• $\mu = 0.93$ (coefficient of friction),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $d = 13.1 \, \text{m}$ (distance over which the sled slides before stopping).

Since the sled comes to a stop, the initial kinetic energy ($KE$) is entirely dissipated by the work done by friction: $KE = W$ The initial kinetic energy is: $KE = \frac{1}{2} m v^2$ Setting $KE = W$ gives: $\frac{1}{2} m v^2 = \mu \cdot m \cdot g \cdot d$ Notice that $m$ cancels out from both sides: $\frac{1}{2} v^2 = \mu \cdot g \cdot d$ Solving for $v$: $v = \sqrt{2 \cdot \mu \cdot g \cdot d}$

Now, plug in the values: $v = \sqrt{2 \cdot 0.93 \cdot 9.8 \cdot 13.1}$

Calculating this: $v \approx \sqrt{2 \cdot 0.93 \cdot 9.8 \cdot 13.1} \approx \sqrt{238.308} \approx 15.4 \, \text{m/s}$

So, the sled was originally moving at approximately 15.4 m/s.

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Related Questions

1. How would the initial speed change if the coefficient of friction was lower?
2. What would be the initial speed if the distance was doubled?
3. How does increasing the mass of the sled affect the initial speed calculation?
4. What additional information would be needed to calculate the time the sled took to stop?
5. How does the gravitational force influence the stopping distance in such scenarios?

Tip: The coefficient of friction indicates how much resistance a surface provides. A higher value means more frictional force for a given normal force.