Problem 76 - University Physics with Modern Physics

Question

The spring of a spring gun has force constant \(k = 400\) N/m and negligible mass. The spring is compressed 6.00 cm, and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so that the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)

Step-by-Step Solution

Answer

(a) 6.93 m/s, (b) 4.90 m/s, (c) Max speed 5.75 m/s at 1.5 cm.

Step 1: Understanding the Problem

We are given a spring gun with a force constant \( k = 400 \) N/m. Initially, the spring is compressed by 6.00 cm (0.06 m). A ball of mass 0.0300 kg is placed in the barrel, which is 6.00 cm long, and is then released. We are to calculate:(a) The speed of the ball ignoring friction.(b) The speed of the ball considering a 6.00 N resisting force.(c) The position of maximum speed along the barrel and that maximum speed considering resisting force.

Step 2: Calculate Initial Spring Potential Energy

When the spring is compressed, potential energy stored in the spring is given by the formula: \( U = \frac{1}{2} k x^2 \).Substitute \( k = 400 \) N/m and \( x = 0.06 \) m:\[ U = \frac{1}{2} \times 400 \times (0.06)^2 = 0.72 \, \text{J} \]

Step 3: Energy Conservation for Part (a)

Assuming no friction, all the spring potential energy converts into kinetic energy of the ball: \( K = \frac{1}{2} m v^2 \).Equate the initial potential energy to kinetic energy and solve for \( v \):\[ \frac{1}{2} \times 0.0300 \times v^2 = 0.72 \]\[ v^2 = \frac{0.72 \times 2}{0.0300} = 48 \]\[ v = \sqrt{48} \approx 6.93 \, \text{m/s} \]

Step 4: Calculate the Effect of Resisting Force for Part (b)

For part (b), consider the work done by the resisting force: \( W = F \cdot d \) where \( F = 6.00 \) N and \( d = 0.06 \) m.\[ W = 6.00 \times 0.06 = 0.36 \, \text{J} \]The kinetic energy of the ball is reduced by this work:\[ 0.72 \, \text{J} - 0.36 \, \text{J} = 0.36 \, \text{J} \]Set the reduced energy equal to kinetic energy and solve for \( v \):\[ \frac{1}{2} \times 0.0300 \times v^2 = 0.36 \]\[ v^2 = \frac{0.36 \times 2}{0.0300} = 24 \]\[ v = \sqrt{24} \approx 4.90 \, \text{m/s} \]

Step 5: Find Maximum Speed and Position for Part (c)

The speed of the ball is maximum when the force due to the spring equals the resisting force. The spring force is given by \( F_s = kx \).Set \( F_s = 6.00 \) N (resisting force):\[ 400x = 6.00 \]\[ x = 0.015 \, \text{m} \] (or 1.5 cm from start)Energy at \( x \):\[ U = \frac{1}{2} \times 400 \times (0.015)^2 = 0.045 \, \text{J} \]Remaining energy is kinetic:\[ 0.72 \, \text{J} - 0.045 \, \text{J} - (0.06 \, \text{m} - 0.015 \, \text{m}) \times 6.00 = 0.495 \, \text{J} \]Solve for maximum speed \( v_m \):\[ \frac{1}{2} \times 0.0300 \times v_m^2 = 0.495 \]\[ v_m^2 = \frac{0.495 \times 2}{0.0300} = 33 \]\[ v_m = \sqrt{33} \approx 5.75 \, \text{m/s} \]

Key Concepts

Spring Potential Energy

Spring potential energy is the energy stored in a spring when it is compressed or stretched from its rest position. In our exercise, the spring is initially compressed 6.00 cm. This compression transforms mechanical energy into potential energy. The formula to calculate spring potential energy is:
\[ U = \frac{1}{2} k x^2 \]
Where:

• \( k \) is the spring constant (400 N/m in this case)
• \( x \) is the displacement from the equilibrium position (0.06 m)

Plugging in the numbers, the spring's potential energy is:
\[ U = \frac{1}{2} \times 400 \times (0.06)^2 = 0.72 \text{ J} \]
This stored energy will convert into kinetic energy as it releases, moving the ball down the gun barrel.

Kinetic Energy

Once the spring is released, its potential energy converts into the kinetic energy of the ball. Kinetic energy is the energy of motion, calculated by:
\[ K = \frac{1}{2} m v^2 \]
Where:

• \( m \) is the mass of the ball (0.0300 kg)
• \( v \) is the velocity we are solving for

For part (a), ignoring any resisting force, the entire 0.72 J of potential energy converts to kinetic energy:
\[ \frac{1}{2} \times 0.0300 \times v^2 = 0.72 \]
Solving for \( v \):
\[ v^2 = \frac{0.72 \times 2}{0.0300} = 48 \]
\[ v = \sqrt{48} \approx 6.93 \text{ m/s} \]
This result shows the speed of the ball as it leaves the barrel with maximum conversion efficiency from potential to kinetic energy.

Resisting Force

In part (b) of the exercise, we're introduced to a resisting force, which is essentially an opposing force that reduces the ball's final speed. We are given a constant resisting force of 6.00 N against the ball as it moves through the barrel. This force performs work against the ball's motion, calculated as:
\[ W = F \cdot d \]
Where:

• \( F \) is the force magnitude (6.00 N)
• \( d \) is the distance over which the force acts (0.06 m)

The resulting work is:
\[ W = 6.00 \times 0.06 = 0.36 \text{ J} \]
This work reduces the total kinetic energy available:
\[ 0.72 \text{ J} - 0.36 \text{ J} = 0.36 \text{ J} \]
Solving for the new speed \( v \) under this resisting force:
\[ \frac{1}{2} \times 0.0300 \times v^2 = 0.36 \]
\[ v^2 = \frac{0.36 \times 2}{0.0300} = 24 \]
\[ v = \sqrt{24} \approx 4.90 \text{ m/s} \]
This shows the impact of a resisting force on the energy transfer, slowing down the ball.