Question

When a \(99.5-\mathrm{g}\) slice of bread is inserted into a toaster, the toaster's ejection spring is compressed by \(7.50 \mathrm{~cm}\). When the toaster ejects the toasted slice, the slice reaches a height \(3.0 \mathrm{~cm}\) above its starting position. What is the average force that the ejection spring exerts on the toast? What is the time over which the ejection spring pushes on the toast?

Short answer

Short Answer: The average force exerted by the ejection spring on the toast is approximately \(14.055 \mathrm{N}\), and the time over which the ejection spring pushes on the toast is approximately \(0.02224 \mathrm{s}\).

Step-by-step solution

Calculating the work done by the ejection spring

To find the work done by the ejection spring on the toast, we can use Hooke's law, which states that the force exerted by a spring is proportional to its displacement from its equilibrium position. The formula for Hooke's law is: \(F = -kx\) where \(F\) represents the force, \(k\) represents the spring constant, and \(x\) represents the displacement from the equilibrium position. We don't have information about the spring constant, but we can use the work done by the spring (using the work-energy theorem) to find the spring force exerted on the toast. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy, which can be written as: \(W = \Delta KE\) In this case, the toast starts at rest and ends at rest (at the highest point it reaches), so its initial and final kinetic energies are both zero. Thus, we have: \(W = \frac{1}{2} kx^2\) Now, we need to relate the work done by the spring to the potential energy gained by the toast. Since the toast reaches a height of \(3.0 \mathrm{~cm}\) above its initial position, it gains gravitational potential energy, which can be written as: \(\Delta PE = mgh\) where \(\Delta PE\) represents the change in potential energy, \(m\) is the mass of the toast, \(g\) is the acceleration due to gravity, and \(h\) is the height gained. Now, we can set the work done by the spring equal to the change in potential energy: \(\frac{1}{2} kx^2 = mgh\)

Solving for the average force exerted by the ejection spring

To find the average force exerted by the ejection spring, we can solve the equation we derived in Step 1 for the spring constant \(k\). First, we can plug in the given values for the mass (\(m = 99.5 \mathrm{~g}\)), the height (\(h = 3.0 \mathrm{~cm}\)), and the spring compression (\(x = 7.5 \mathrm{~cm}\)), and noting that \(g = 9.81 \mathrm{~m/s^2}\). \(\frac{1}{2} k(0.075 \mathrm{~m})^2 = (0.0995 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(0.03 \mathrm{~m})\) Now, we can solve for \(k\): \(k = \frac{2 \times (0.0995 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(0.03 \mathrm{~m})}{(0.075 \mathrm{~m})^2} \approx 187.40 \mathrm{N/m}\) With the spring constant \(k\) found, we can now calculate the average force exerted by the ejection spring using Hooke's law: \(F = kx = (187.40 \mathrm{N/m})(0.075 \mathrm{~m}) \approx 14.055 \mathrm{N}\) So, the average force exerted by the ejection spring on the toast is approximately \(14.055 \mathrm{N}\).

Calculating the time over which the ejection spring pushes on the toast

To calculate the time over which the ejection spring pushes on the toast, we will again use the work-energy theorem and the kinematic equations. Since the toast starts at rest and reaches a height of \(3.0 \mathrm{~cm}\), we can write a kinematic equation for its vertical motion: \(v^2 = v_0^2 + 2gh\) where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(h\) is the height. Since the initial velocity is 0, we can use the final velocity: \(v^2 = 2gh\) The toast's final velocity can also be found using the average force exerted by the spring and the toast's mass: \(v = \frac{Ft}{m}\) where \(t\) is the time over which the ejection spring pushes on the toast. Now, we can set these two expressions for the final velocity equal to each other: \(\frac{Ft}{m} = \sqrt{2gh}\) Solving for the time \(t\), we have: \(t = \frac{m \sqrt{2gh}}{F} = \frac{(0.0995 \mathrm{~kg})\sqrt{2(9.81 \mathrm{~m/s^2})(0.03 \mathrm{~m})}}{14.055 \mathrm{N}} \approx 0.02224 \mathrm{s}\) Thus, the time over which the ejection spring pushes on the toast is approximately \(0.02224 \mathrm{s}\).

Key concepts

Understanding Hooke's Law

Hooke's Law is fundamental when dealing with springs and elasticity. According to this principle, the force required to compress or extend a spring is directly proportional to the distance it's stretched or compressed. This relationship can be represented by the equation
\( F = -kx \),
where \( F \) is the force applied to the spring, \( k \) is the spring constant, which measures the stiffness of the spring, and \( x \) is the displacement from the equilibrium position.

In the context of our exercise, Hooke's Law helps us understand the force the toaster's ejection spring applies. To find this force, we first calculated the spring constant by equating the spring's work (equivalent to the mechanical energy imparted to the toast) with the toast's gravitational potential energy at its peak height. It is crucial to note that the force is considered average because the force exerted by a spring changes as it decompresses, being highest at maximum compression and zero at the equilibrium position.

• The formula integrates the nature of the spring and the physical displacement it experiences.
• Understanding this law is important for solving a variety of problems in physics and engineering where spring mechanics are involved.
• For students, visualizing the spring compressing and decompressing can aid in grasping why the force changes with displacement.

Gravitational Potential Energy and Work

Gravitational potential energy (\( PE_g \) is another key concept and is the energy held by an object because of its position relative to a gravitational field. It is calculated using the formula:
\( PE_g = mgh \),
where \( m \) is the object's mass, \( g \) is the acceleration due to gravity (approximately \(9.81 \text{m/s}^2\) on Earth), and \( h \) is the height above the reference point.

Relating it to our exercise, as the toast pops up, it gains gravitational potential energy, reaching a maximum when the toast comes to a temporary stop at the peak of its trajectory. This increase in potential energy is exactly equal to the work done by the spring on the toast, illustrating the work-energy theorem. This theorem states that the work done by all forces acting on an object results in a change in the object's kinetic energy. Since the toast begins and ends at rest, all work done by the spring is converted into potential energy.

• This concept is vital in understanding energy conservation in systems influenced by gravity.
• Students should remember that \( PE_g \) only depends on the vertical distance and not the path taken by the object.
• Visualizing energy transfer from the spring (mechanical) to the toast (gravitational) can help cement this concept.

Dynamics of Motion: Kinematic Equations

Kinematic equations describe the motion of objects without considering the forces that cause the motion. They relate variables like displacement, initial and final velocities, acceleration, and time. These equations are essential for analyzing motion in physics.

In the current exercise, the kinematic equation \( v^2 = v_0^2 + 2gh \) is utilized, where \( v \) is the final velocity, \( v_0 \) is the initial velocity, and \( h \) is the height. Since the toast leaves the toaster at rest and is momentarily at rest at the peak, the kinematic equation simplifies to \( v^2 = 2gh \), making it easier to find the final velocity. This final velocity, when set equal to the velocity derived from Newton's second law (\( v = Ft/m \) for the force over a time period, provides a route to calculate the time the spring acts on the toast.

• These equations are crucial tools for predicting subsequent positions and velocities of moving objects.
• Understanding how time, velocity, and acceleration interplay is pivotal for solving motion-related problems in physics.
• Always check initial conditions, such as whether the initial velocity is zero, as these simplify calculations significantly.