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Chapter 8: Problem 18

Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice. (a) If the heavier skater travels at 1.50 m/s, how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this energy come from?

Short Answer

Expert verified
(a) 1.74 m/s; (b) 180.00 J from muscle work.

Step by step solution

01

Understand the Problem

We have two skaters of different weights pushing off each other on frictionless ice. We need to find the speed of the lighter skater, given the speed of the heavier skater, and calculate the kinetic energy involved in the maneuver.
02

Apply Conservation of Momentum

Since the skaters push off against each other on frictionless ice, momentum is conserved. The momentum of the heavier skater is mass times velocity. To find the mass, recall weight = mass times gravity. Let's find both masses:\( m_1 = \frac{625}{9.8} \approx 63.78 \text{ kg} \) for the lighter skater,\( m_2 = \frac{725}{9.8} \approx 73.98 \text{ kg} \) for the heavier skater.Using the conservation of momentum:\( m_1 \cdot v_1 = m_2 \cdot v_2 \)\( 63.78 \cdot v_1 = 73.98 \cdot 1.50 \) where \( v_2 = 1.50 \) m/s.
03

Solve for the Velocity of the Lighter Skater

Let's solve for \( v_1 \):\[ v_1 = \frac{73.98 \times 1.50}{63.78} \approx 1.74 \text{ m/s} \]The lighter skater will travel at approximately 1.74 m/s.
04

Calculate Kinetic Energy Created

The kinetic energy of each skater can be calculated using the formula \( KE = \frac{1}{2}mv^2 \).For the heavier skater:\( KE_2 = \frac{1}{2} \times 73.98 \times (1.50)^2 \approx 83.48 \text{ J} \).For the lighter skater:\( KE_1 = \frac{1}{2} \times 63.78 \times (1.74)^2 \approx 96.52 \text{ J} \).The total kinetic energy created during the maneuver is:\( KE_{total} = KE_1 + KE_2 = 83.48 + 96.52 = 180.00 \text{ J} \).
05

Understand the Source of Energy

The energy created comes from the work done by the skaters against each other. As they push off, they convert the energy exerted during the push into kinetic energy. This energy comes from the muscle power of the skaters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In the exercise, the skaters each gain kinetic energy as they move across the ice. The heavier skater, moving at 1.50 m/s, creates kinetic energy of about 83.48 J. Meanwhile, the lighter skater, moving at approximately 1.74 m/s, creates 96.52 J. Together, they produce a total of 180 J of kinetic energy. This energy is "created" from the work that the skaters did against each other when they pushed off. Understanding kinetic energy is crucial for solving various physics problems involving motion.
Physics Problem Solving
When faced with a physics problem, it is essential to break it down into smaller, manageable parts.
  • Identify what information is given and what needs to be found.
  • Apply relevant laws and formulas.
  • Solve the equations step by step while verifying units and dimensions.
In this problem, we needed to determine the lighter skater's speed and the kinetic energy generated. We first identified the weights and speeds given for the skaters, then used the conservation of momentum to find the unknown speed. Subsequently, we applied the kinetic energy formula to compute energy values. Always make sure to understand the problem fully before attempting to solve it.
Frictionless Surface
The concept of a frictionless surface plays an important role in physics problems as it simplifies the scenario being analyzed. On such a surface, like the ice in this problem, there are no forces opposing motion. This means the only forces to consider are those applied by the skaters themselves.
  • No energy is lost to friction, which helps in preserving the consistency of momentum calculations.
  • It aids in illustrating the principle of conservation processes like momentum and energy.
In this skater scenario, it is this frictionless condition that permits the skaters to smoothly push away from each other without loss of energy to the environment, allowing us to simply apply the conservation of momentum.
Newton's Laws
Newton's Laws of Motion form the basis for understanding this problem. The third law, in particular, is highly relevant: "For every action, there is an equal and opposite reaction." As the skaters push off against each other, both exert forces that are equal in magnitude but opposite in direction. This interaction results in the conservation of momentum. Furthermore, because the net external force acting on the skating system is zero (thanks to the frictionless surface), the total momentum before and after they push off remains unchanged. Newton's Laws, therefore, provide a fundamental explanation of why each skater has a certain velocity and how the energy transfer occurs during their maneuver.

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Most popular questions from this chapter

A 10.0-g marble slides to the left at a speed of 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head- on, elastic collision with a larger 30.0-g marble sliding to the right at a speed of 0.200 m/s (\(\textbf{Fig. E8.48}\)). (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all motion is along a line.) (b) Calculate the \(change\) \(in\) \(momentum\) (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the \(change\) \(in\) \(kinetic\) \(energy\) (the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

A 5.00-g bullet is shot \(through\) a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s.

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring \((\textbf{Fig. P8.79})\). The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

Two cars collide at an intersection. Car \(A\), with a mass of 2000 kg, is going from west to east, while car \(B\), of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65\(^\circ\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (\(\textbf{Fig. E8.55}\)). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90\(^\circ\) to make the entire part horizontal?
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