Problem 63 - College Physics

Question

\(\bullet\) A movie stuntman (mass 80.0 \(\mathrm{kg}\) ) stands on a window ledge 5.0 \(\mathrm{m}\) above the floor (Fig. \(8.45 ) .\) Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 \(\mathrm{kg}\) ), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 \(\mathrm{m} .\) He releases the rope just as he reaches the villain.) (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is \(\mu_{k}=0.250,\) how far do they slide?

Step-by-Step Solution

Answer

The foes slide across the floor with an initial speed of 5.27 m/s and slide 5.67 m before stopping.

Step 1: Calculate Potential Energy of Stuntman

The potential energy (PE) of the stuntman at a height of 5.0 m is given by \( PE = mgh \), where \( m = 80.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 5.0 \, \text{m} \). Thus, \( PE = 80.0 \times 9.81 \times 5.0 = 3924 \, \text{J} \).

Step 2: Convert Potential Energy to Kinetic Energy

As the stuntman swings down, all his potential energy is converted to kinetic energy (KE) at the bottom. Thus, \( KE = 3924 \, \text{J} \). Use the formula \( KE = \frac{1}{2}mv^2 \) to find the speed just before the collision with the villain: \( 3924 = \frac{1}{2} \times 80.0 \times v^2 \). Solving for \( v \), we get \( v = \sqrt{\frac{2 \times 3924}{80.0}} \approx 9.89 \, \text{m/s} \).

Step 3: Apply Conservation of Momentum for Collision

Just before the stuntman grabs the villain, their combined system's initial momentum is only that of the stuntman. Apply conservation of momentum: \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \). The villain is initially at rest so \( v_2 = 0 \). Calculate \( (80.0 \times 9.89) + (70.0 \times 0) = (80.0 + 70.0)v_f \). Solving for \( v_f \), \( v_f = \frac{80.0 \times 9.89}{150.0} \approx 5.27 \, \text{m/s} \).

Step 4: Calculate Friction Force and Deceleration

The kinetic friction force \( f_k \) acting on them is \( f_k = \mu_k (m_1 + m_2)g \) where \( \mu_k = 0.250 \). Thus, \( f_k = 0.250 \times 150.0 \times 9.81 = 367.875 \, \text{N} \). The deceleration \( a \) due to friction is \( a = \frac{f_k}{m_1 + m_2} = \frac{367.875}{150.0} = 2.4525 \, \text{m/s}^2 \).

Step 5: Calculate Sliding Distance using Kinematic Equation

Use the kinematic equation \( v_f^2 = v_i^2 + 2a d \), where \( v_f = 0 \), \( v_i = 5.27 \, \text{m/s} \), and \( a = -2.4525 \, \text{m/s}^2 \). Solve for \( d \): \( 0 = (5.27)^2 - 2 \times 2.4525 \times d \). Thus, \( d = \frac{(5.27)^2}{2 \times 2.4525} \approx 5.67 \, \text{m} \).

Key Concepts

Kinetic Energy

Kinetic energy is a critical concept when analyzing motion, especially in collision physics. It represents the energy an object possesses due to its motion and is given by the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity.

In our example, the stuntman initially has potential energy due to his height which is completely converted to kinetic energy as he swings downwards. This means before reaching the villain, all of his stored energy has been transformed into motion energy, captured in the speed calculated just before the collision, approximately \( 9.89 \text{ m/s} \).

It's important to visualize kinetic energy as the actual energy of motion that gets transferred or shared upon interaction with another object, like when the stuntman meets the villain.

Momentum Conservation

Momentum conservation is a fundamental principle in physics that states the total momentum of a closed system is constant if no external forces act on it. In simpler terms, for any collision or interaction where no outside push or pull exists, the motion aspect stays unchanged.

When the stuntman swings down and grabs the villain, the conservation of momentum comes into play. Initially, only the stuntman has momentum, as the villain is standing still. By using the principle \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \), you compute the shared momentum post-collision. Here, the combined speed is found to be about \( 5.27 \text{ m/s} \), demonstrating how momentum transfers between the two bodies as they begin to move as one unit.

Friction

Friction is the force that acts against movement, making it necessary to work harder to start and maintain motion. It plays a critical role in understanding how objects slow down or come to a stop. Friction is measured with the coefficient of friction \( \mu\), which varies based on the surfaces in contact.

After the stuntman and villain collide, they slide across the floor where friction is the main resistance slowing them. Here, the kinetic friction force (\( f_k \)) is calculated using \( f_k = \mu_k(m_1 + m_2)g \), showing how friction force depends on both the coefficient \( \mu_k = 0.250 \) and their combined weight. This force leads to deceleration, letting us compute how far they travel before stopping.

Potential Energy

Potential energy is stored energy due to an object's position or configuration. In the context of height, like the stuntman standing on a ledge, it can be viewed as the energy ready for conversion into other energy forms, such as motion. This form of energy is calculable using \( PE = mgh \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height.

The stuntman's potential energy, given by his elevation of \( 5 \text{ m} \), is \( 3924 \text{ J} \). This value represents the total energy available for doing work when he descends. By the time he reaches the rope's lowest point, all potential energy is converted into kinetic energy, perfectly encapsulating energy conservation principles. Understanding potential energy helps us track energy transformation in various physical scenarios, including the initial steps leading to a collision.