A sled is moving at a constant speed when it encounters a patch of rough ground. If the coefficient of friction is 0.93 and the sled slides for 13.1 meters before coming to a halt, then how fast was the sled originally moving? (please provide you answer to 1 decimal place)

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**.

Would you like more details on the steps?

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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.

A sled is moving at a constant speed when it encounters a patch of rough ground. If the coefficient of friction is 0.93 and the sled slides for 13.1 meters before coming to a halt, then how fast was the sled originally moving?

Find out how to calculate the initial speed of a sled that stops after encountering friction over 13.1 meters. Using the work-energy principle, this problem explores key concepts like kinetic energy, friction, and involves the formula v = √(2 × μ × g × d). Ideal for high school students in Grades 10-12.

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