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Refer to Concept Simulation 6.1 at for a review of the concepts with which this problem deals. A \(0.075-\mathrm{kg}\) arrow is fired horizontally. The bowstring exerts an average force of \(65 \mathrm{~N}\) on the arrow over a distance of \(0.90 \mathrm{~m}\). With what speed does the arrow leave the bow?

Short Answer

Expert verified
The arrow leaves the bow at approximately \(39.5\, \text{m/s}\).

Step by step solution

01

Identify Given Values

We are given the mass of the arrow as \(0.075\, \text{kg}\), the force exerted by the bowstring as \(65\, \text{N}\), and the distance over which this force acts as \(0.90\, \text{m}\).
02

Use the Work-Energy Principle

The work done on the arrow is equal to the kinetic energy gained by it. Work done \(W\) can be calculated as \(F \cdot d\), where \(F = 65\, \text{N}\) and \(d = 0.90\, \text{m}\).
03

Calculate Work Done

Calculate the work done on the arrow: \[ W = 65\, \text{N} \times 0.90\, \text{m} = 58.5\, \text{J} \]
04

Relate Work to Kinetic Energy

According to the work-energy principle, the work done on the arrow is equal to its kinetic energy: \[ W = \frac{1}{2}mv^2 \] where \(m = 0.075\, \text{kg}\) and \(W = 58.5\, \text{J}\).
05

Solve for Velocity

Substitute the values into the kinetic energy formula: \[ 58.5 = \frac{1}{2}\times 0.075 \times v^2 \] Solve for \(v\). Multiply both sides by 2 and divide by \(0.075\): \[ v^2 = \frac{2 \times 58.5}{0.075} \] \[ v^2 = 1560 \] \[ v = \sqrt{1560} \approx 39.5\, \text{m/s} \]
06

Conclude

The speed of the arrow as it leaves the bow is approximately \(39.5\, \text{m/s}\).

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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. If you have ever seen a moving car or a running person, you've witnessed kinetic energy in action! The amount of kinetic energy an object has depends on two factors: its mass and its velocity. Mathematically, kinetic energy is defined by the formula: \[ KE = \frac{1}{2} m v^2 \] where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity. It's important to note that kinetic energy grows with the square of the velocity, meaning a faster object carries much more energy. In our arrow example, understanding kinetic energy explains why the arrow gains a certain speed - the bow transforms potential energy into this kinetic form as it releases the arrow.
Work Done
When we talk about work in physics, we mean the energy transferred to an object via a force that causes displacement. It is different from the everyday use of the word 'work.' In our scenario, the bowstring exerts a force on the arrow, pushing it and doing 'work' on it.The amount of work done is calculated as: \[ W = F \cdot d \] where \( W \) is work, \( F \) is the force applied, and \( d \) is the distance over which the force is applied. In this scenario, the bowstring exerted a force of 65 N over a distance of 0.90 m, resulting in 58.5 J of work done on the arrow. This value of work directly contributes to the arrow's kinetic energy as it leaves the bow, converting the exerted force into the speed of the arrow.
Force and Motion
Force is a crucial concept in understanding motion. It's the interaction that, when unopposed, changes the motion of an object. Think of it as a push or a pull. The direction and magnitude of a force can determine how and if an object will start moving, stop moving, or change direction. In our example, the force exerted by the bowstring is what sets the arrow in motion. The magnitude of 65 N is substantial for the arrow, and because there's a clear path, this force efficiently accelerates the arrow. This constant force applied over the distance of the bowstring's pull propels the arrow forward with increasing velocity, showcasing how force alters motion.
Mechanics
Mechanics is the branch of physics that deals with the behavior of physical bodies when subjected to forces or displacements. It encompasses various topics, including but not limited to the concepts of force, energy, and motion. Mechanics can be divided into kinematics and dynamics.
  • Kinematics: This area focuses on the motion of objects without considering the forces that cause the motion. It's essentially the "how" of motion - describing speed, velocity, and acceleration.
  • Dynamics: This area considers the "why" - addressing the influences of forces and torques and the nature of interactions. It includes studying Newton's Laws of Motion, which are pivotal in understanding interactions like those seen in our arrow problem.
For our arrow problem, mechanics aids in understanding how the arrow's speed is achieved through the dynamics of the bowstring's force combined with the kinematic motion as the arrow flies through the air. Thus, mechanics gives us a comprehensive view of the system at play.

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Most popular questions from this chapter

The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of \(4.0 \mathrm{~m} / \mathrm{s}\), what is the largest value that the radius \(r\) can have if the car is to remain in contact with the circular track at all times?

During a tug-of-war, team A pulls on team B by applying a force of \(1100 \mathrm{~N}\) to the rope between them. How much work does team A do if they pull team B toward them a distance of \(2.0 \mathrm{~m} ?\)

Refer to Interactive Solution \(6.51\) for a review of the approach taken in problems such as this one. A \(67.0\) -kg person jumps from rest off a \(3.00\) -m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest \(1.10 \mathrm{~m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

The (nonconservative) force propelling \(1.50 \times 10^{3}-\mathrm{kg}\) a car up a mountain road does \(4.70 \times 10^{6} \mathrm{~J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{~m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{~m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

A \(55.0-\mathrm{kg}\) skateboarder starts out with a speed of \(1.80 \mathrm{~m} / \mathrm{s} .\) He does \(+80.0 \mathrm{~J}\) of work on himself by pushing with his feet against the ground. In addition, friction does \(-265 \mathrm{~J}\) of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{~m} / \mathrm{s}\). (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

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