A physics professor pretending to be spider man swings from a rope that is 12 meters long starting at an angle of 45° down to the ground then skids to a halt. If the professor has a mass of 80 kg and a coefficient of friction with the ground of 0.90, how far does he slide?

To solve this problem, we need to approach it step by step. Here's how we can break it down:

### Step 1: Determine the velocity of the professor at the moment he reaches the ground.
The professor is swinging down from a rope, which means his potential energy at the starting point is converted into kinetic energy at the lowest point of the swing.

The initial potential energy is given by:

\[
PE = mgh
\]

where:
- \(m = 80 \, \text{kg}\) (mass of the professor),
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity),
- \(h = L - L \cos \theta\) (height fallen, where \(L = 12 \, \text{m}\) is the length of the rope, and \(\theta = 45^\circ\)).

First, calculate the height:

\[
h = 12 - 12 \cos 45^\circ
\]
\[
h = 12 - 12 \times \frac{\sqrt{2}}{2} = 12 - 12 \times 0.707 = 12 - 8.49 = 3.51 \, \text{m}
\]

Now calculate the potential energy:

\[
PE = 80 \times 9.8 \times 3.51 = 2742.24 \, \text{J}
\]

Since all the potential energy converts to kinetic energy at the bottom of the swing, we can use:

\[
KE = \frac{1}{2}mv^2
\]

Setting \(PE = KE\):

\[
2742.24 = \frac{1}{2} \times 80 \times v^2
\]
\[
2742.24 = 40v^2
\]
\[
v^2 = \frac{2742.24}{40} = 68.56
\]
\[
v = \sqrt{68.56} \approx 8.28 \, \text{m/s}
\]

So, the professor's velocity at the lowest point of the swing is approximately **8.28 m/s**.

### Step 2: Calculate the frictional force.
Once the professor reaches the ground, he begins to slide and experiences a frictional force. The force of friction is given by:

\[
f_{\text{friction}} = \mu N
\]

where:
- \(\mu = 0.90\) is the coefficient of friction,
- \(N = mg\) is the normal force, which in this case equals the weight of the professor.

\[
N = 80 \times 9.8 = 784 \, \text{N}
\]
\[
f_{\text{friction}} = 0.90 \times 784 = 705.6 \, \text{N}
\]

### Step 3: Calculate the deceleration due to friction.
Using Newton's second law, the deceleration \(a\) due to friction is:

\[
f_{\text{friction}} = ma
\]

\[
705.6 = 80 \times a
\]

\[
a = \frac{705.6}{80} = 8.82 \, \text{m/s}^2
\]

### Step 4: Calculate the distance the professor slides.
To calculate the distance the professor slides, we can use the following kinematic equation:

\[
v^2 = u^2 + 2ad
\]

where:
- \(v = 0\) m/s is the final velocity (since he comes to a halt),
- \(u = 8.28\) m/s is the initial velocity,
- \(a = -8.82\) m/s² is the acceleration (negative because it's deceleration),
- \(d\) is the distance traveled.

Rearrange the equation to solve for \(d\):

\[
0 = (8.28)^2 + 2(-8.82)d
\]
\[
0 = 68.56 - 17.64d
\]
\[
17.64d = 68.56
\]
\[
d = \frac{68.56}{17.64} \approx 3.89 \, \text{m}
\]

### Final Answer:
The professor slides approximately **3.89 meters**.

---

Would you like further details or clarification on any part of the solution? 

Here are some related questions you might find interesting:
1. How does the angle of the swing affect the velocity at the lowest point?
2. What would happen if the coefficient of friction were lower?
3. How would the distance change if the professor had a different mass?
4. What if the rope were longer? How would it affect the motion?
5. How does the conservation of mechanical energy apply to this scenario?

**Tip**: When analyzing problems involving swinging or motion, always consider energy conservation principles and forces acting in the direction of motion to solve for unknowns like velocity and distance.

Learn how to calculate the sliding distance of a professor swinging from a 12-meter rope at a 45° angle. This problem covers energy conservation, kinematic equations, and frictional force for an object with an 80 kg mass and a coefficient of friction of 0.90. Suitable for Grades 11-12.

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