Math Problem Statement

Someone with a mass of 86kg swings from a rope that is 14m long at angle of 45 degrees. The coefficient of friction is 0.5. How far did they slide?

Solution

To determine how far the person slid, we first need to analyze the energy involved in the swing and friction.

Key points:

  • Mass (m): 86 kg
  • Length of rope (L): 14 m
  • Angle of swing (θ): 45°
  • Coefficient of friction (μ): 0.5
  • We assume the sliding happens when the person lands after swinging.

Step 1: Determine the potential energy at the highest point of the swing

At the start of the swing, the person is at a height hh above the lowest point of the swing. We calculate the vertical height using trigonometry: h=LLcos(θ)h = L - L \cos(θ) h=1414cos(45)=1414×0.7071=149.89944.1mh = 14 - 14 \cos(45^\circ) = 14 - 14 \times 0.7071 = 14 - 9.8994 \approx 4.1 \, \text{m}

Step 2: Calculate the gravitational potential energy

The gravitational potential energy (PE) at the highest point is given by: PE=mghPE = mgh Where:

  • m=86kgm = 86 \, \text{kg} (mass),
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2 (acceleration due to gravity),
  • h=4.1mh = 4.1 \, \text{m}.

PE=86×9.8×4.13454.28JPE = 86 \times 9.8 \times 4.1 \approx 3454.28 \, \text{J}

Step 3: Energy converted to kinetic energy

As the person swings down, all this potential energy is converted into kinetic energy (ignoring air resistance) at the lowest point of the swing: KE=12mv2KE = \frac{1}{2} mv^2 So, KE=3454.28JKE = 3454.28 \, \text{J}.

Step 4: Work done by friction (to stop the person)

When the person lands, they slide along the ground. The frictional force does work to stop the motion. The work done by friction WfW_f is given by: Wf=μNdW_f = \mu \cdot N \cdot d Where:

  • μ=0.5\mu = 0.5 (coefficient of friction),
  • N=mgN = mg is the normal force, and
  • dd is the distance slid.

The normal force NN is simply the person’s weight: N=mg=86×9.8=842.8NN = mg = 86 \times 9.8 = 842.8 \, \text{N}

Thus, the work done by friction becomes: Wf=0.5×842.8×d=421.4×dW_f = 0.5 \times 842.8 \times d = 421.4 \times d

Step 5: Set the work done by friction equal to the kinetic energy

The work done by friction will equal the kinetic energy (since friction brings the person to rest): 421.4×d=3454.28421.4 \times d = 3454.28

Solving for dd: d=3454.28421.48.2md = \frac{3454.28}{421.4} \approx 8.2 \, \text{m}

Final Answer:

The person slid approximately 8.2 meters.


Would you like further details or have any other questions?

Here are 5 related questions to explore further:

  1. What happens if the coefficient of friction were increased to 0.6?
  2. How would a change in the swing angle affect the distance slid?
  3. What if the mass of the person was different, say 75 kg, how would that impact the result?
  4. How does air resistance affect the energy conversion during the swing?
  5. How can we calculate the velocity at the lowest point of the swing?

Tip: When solving problems involving friction, always remember to account for the normal force, which is the weight of the object if it's sliding horizontally.

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Math Problem Analysis

Mathematical Concepts

Energy Conservation
Kinetic Energy
Potential Energy
Work and Friction
Trigonometry

Formulas

Potential Energy: PE = mgh
Kinetic Energy: KE = (1/2)mv^2
Work Done by Friction: W_f = μNd
Height Calculation: h = L - Lcos(θ)

Theorems

Conservation of Energy

Suitable Grade Level

Grades 10-12