Question

If it takes an average force of \(5.50 \mathrm{~N}\) to push a \(4.50-\mathrm{g}\) dart \(6.00 \mathrm{~cm}\) into a dart gun, assuming that the barrel is frictionless, how fast will the dart exit the gun?

Short answer

Answer: The final velocity of the dart when it exits the dart gun is approximately 12.11 m/s.

Step-by-step solution

Write down the given information and relevant formulas.

We are given: - Average force, \(F_{avg} = 5.50\ \text{N}\) - Mass of the dart, \(m = 4.50\ \text{g}\) - Distance force is applied, \(d = 6.00\ \text{cm}\) We need to find the final velocity, \(v_f\). Relevant formulas: - Work-energy theorem: \(W = \Delta KE\) - Work done by force: \(W = F_{avg} \cdot d\) - Change in kinetic energy: \(\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)\) - Initially, the dart is at rest, so \(v_i = 0.\)

Convert quantities to SI units.

We need to ensure all quantities are in SI units before proceeding. Mass of the dart in kg: \(m = 4.50\ \text{g} \cdot \frac{1\ \text{kg}}{1000\ \text{g}} = 0.0045\ \text{kg}\) Distance force is applied in meters: \(d = 6.00\ \text{cm} \cdot \frac{1\ \text{m}}{100\ \text{cm}} = 0.0600\ \text{m}\) Now we have: - \(F_{avg} = 5.50\ \text{N}\) - \(m = 0.0045\ \text{kg}\) - \(d = 0.0600\ \text{m}\)

Calculate the work done by the force.

Use the formula for work done by force: \(W = F_{avg} \cdot d\) \(W = (5.50\ \text{N})(0.0600\ \text{m}) = 0.330\ \text{J}\) The work done by the force is \(0.330\ \text{J}\).

Use the work-energy theorem to find the change in kinetic energy.

Apply the work-energy theorem: \(W = \Delta KE\) Since the initial velocity is \(0\ \text{m/s}\), the change in kinetic energy formula simplifies to: \(\Delta KE = \frac{1}{2}mv_f^2\) Now we can set up the equation using the work-energy theorem: \(0.330\ \text{J} = \frac{1}{2}(0.0045\ \text{kg})v_f^2\)

Solve for the final velocity.

Rearrange the equation to solve for \(v_f\): \(v_f^2 = \frac{2 \cdot 0.330\ \text{J}}{0.0045\ \text{kg}}\) \(v_f^2 = 146.67\) \(v_f = \sqrt{146.67}\) \(v_f \approx 12.11\ \text{m/s}\) The dart will exit the gun at approximately \(12.11\ \text{m/s}\).

Key concepts

Work-Energy Theorem

Understanding the work-energy theorem is crucial when analyzing the motion of objects in physics. Essentially, this theorem connects the work done by all forces acting on an object to the change in its kinetic energy. In simpler terms, it tells us that the total work done on an object will result in a corresponding change in its kinetic energy. For instance, when you apply force to push a dart into a dart gun, you are doing work on the dart. The work-energy theorem is mathematically expressed as \( W = \Delta KE \), where \( W \) represents the work done on the object, and \( \Delta KE \) signifies the change in kinetic energy.

In our current example, the work done to push the dart is converted entirely into the dart's kinetic energy since no other forces, like friction, are acting on it. This scenario perfectly illustrates how the work-energy theorem can be applied to determine the speed of an object after a certain amount of work is done on it.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It's an important concept in physics because it helps to understand how the velocity of an object affects its energy. An object's kinetic energy can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

In the context of our problem, the dart's kinetic energy is what we're solving for because we want to know how fast the dart will exit the gun. By rearranging the work-energy theorem, we directly calculate the dart's final kinetic energy, which then allows us to find its final velocity upon exiting the dart gun. This exercise demonstrates how the conservation of energy principle is used in practice: the energy we put into the system (work done) becomes the dart's energy of motion (kinetic energy).

Force and Motion

Force and motion are fundamental concepts within physics, and they are intimately tied to understanding how objects move and interact. Newton's laws, particularly the second law, which states that \( F = ma \) (force equals mass times acceleration), form the foundation of these concepts. When a force is applied to an object, it can cause the object to accelerate, change direction, or alter its velocity.

In our dart-gun scenario, the force applied to push the dart is what initiates the dart's motion. The mass and the applied force are used to calculate the dart's kinetic energy and thus determine its resulting motion or final velocity. What's interesting here is that although an object's acceleration is directly proportional to the force applied, the final velocity is not. It's a function of the work done, which incorporates both force and displacement, giving us a clearer picture of the dart’s energy change and ultimately its speed as it exits the gun.