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In the winter sport of curling, two teams alternate sliding \(20 \mathrm{kg}\) stones on an icy surface in an attempt to end up with the stone closest to the center of a target painted on the ice. During one turn, a player releases a stone that travels \(27.9 \mathrm{m}\) before coming to rest. The friction force acting on the stone is \(2.0 \mathrm{N}\). What was the speed of the stone when the player released it?

Short Answer

Expert verified
The initial speed of the stone when the player released it was approximately \(4.74 \mathrm{m/s}\).

Step by step solution

01

Identify the known variables

The mass (\(m\)) of the stone is given as \(20 \mathrm{kg}\), the force of friction (\(F\)) is \(2.0 \mathrm{N}\), and the distance (\(d\)) it traveled is \(27.9 \mathrm{m}\) before it came to rest.
02

Apply the work-energy theorem

The work-energy theorem states that work done on an object is equal to the change in its kinetic energy. Given that the final kinetic energy of the stone is zero (since it comes to rest), we can write the work done by friction as \[W = \frac{1}{2}mv^2\] where \(W\) is the work done, \(m\) is the mass, and \(v\) is the velocity.
03

Calculate the work done by friction

The work done by friction is the force of friction times the distance the stone traveled, hence \(W = Fd\). Substituting the given values, \[W = 2.0 \mathrm{N} \times 27.9 \mathrm{m} = 55.8 \mathrm{J}\]
04

Solve the equation for velocity

Substitute the calculated work and the given mass into the equation for work, then solve for velocity (\(v\)). \[\frac{1}{2} \times 20 \mathrm{kg} \times v^2 = 55.8 \mathrm{J}\] Solving for \(v\), we obtain \(v = \sqrt{\frac{2 \times 55.8 \mathrm{J}}{20 \mathrm{kg}}} = 4.74 \mathrm{m/s}\]

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Key Concepts

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

Understanding the Work-Energy Theorem
The work-energy theorem is a fundamental concept in physics that connects the dots between the work done on an object and the object's kinetic energy. It tells us that the work done by all the forces acting on an object results in a change in the object's kinetic energy. Think of it like a trade-off between effort (work) and motion (kinetic energy).

In our curling stone example, the only work being done on the stone comes from friction force. When the stone is sliding across the ice, friction acts to slow it down, working against the stone's movement. This friction force removes kinetic energy from the stone, eventually bringing it to a stop. In mathematical terms, the theorem can be expressed as \( W = \Delta K \), where \( W \) is the work done and \( \Delta K \) is the change in kinetic energy.

The key here is recognizing that when the stone stops, all the kinetic energy it had has been transformed into other forms of energy (mostly heat, due to friction), effectively making the final kinetic energy zero. So, the work done by friction has an equal magnitude but opposite sign to the stone's initial kinetic energy, which leads us to our next concept.
Friction Force – The Unseen Hand
Friction force often gets overlooked, yet it plays a crucial role in everyday motions. It's the resistive force that opposes the relative motion between two surfaces in contact. In the context of curling, friction is what eventually brings the stone to rest.

To understand the problem better, envision the moment the player releases the stone. It begins to slide across the ice, and from that moment, friction starts doing its work, sapping the stone's energy. Knowing the force of friction and the distance over which it acts allows us to calculate the work done by friction, with the simple relationship \( W = Fd \), where \( F \) is the consistent friction force and \( d \) is the distance traveled.

For the stone in our problem, the constant friction force is small, only 2.0 N, indicative of the smoothness of the ice. Despite its subtlety, this force is persistent, applying a steady drain on the stone's kinetic energy over the 27.9 meters it travels, effectively transforming the stone's kinetic energy into other forms through the work it has done.
Deciphering Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy, which is directly proportional to both its mass and the square of its velocity. The faster the object moves, or the more it weighs, the more kinetic energy it possesses. For a moving object with mass \( m \) and velocity \( v \) the kinetic energy \( K \) can be calculated utilizing the equation \( K = \frac{1}{2}mv^2 \).

Let's apply this to our curling stone. At the start, the stone is full of kinetic energy, basking in the push given by the player. As it moves along the ice, it gradually loses this energy due to the unyielding friction between the stone and ice. The problem asks us to find the stone's original speed, which we do by setting the stone's kinetic energy equal to the work done by friction, since they are essentially two sides of the same coin.

Finally, by rearranging the kinetic energy formula to solve for the initial velocity, we find the stone's original speed when released. It's a powerful demonstration of how energy transforms from kinetic energy to heat while conserving the overall energy, highlighted by the friction force doing work on the curling stone.

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