How Does a Toaster's Ejection Spring Impact Toast Height?

[the1cyrus]

Homework Statement

When an 86.5-g piece of toast is inserted into a toaster, the toaster's ejection spring is compressed 7.85 cm. When the toaster ejects the toasted slice, the slice reaches a height of 3.9 cm above it's 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. Assume that throughout the ejection process the toast experiences a constant acceleration.

Homework Equations

Not sure. I'm confused about 'assumptions' with the question. Seems broken to me. I approached it using different ways; 1. finding the spring constant of the spring, plugging that into the force formula [f=kx] OR ignoring the fact that the spring is a spring, then comparing forces [setting them equal to each other] and solving for some variable. The first method seems to work, but has given twice the correct answer. I'm not looking for the answer, I'm looking for the methodology.

The Attempt at a Solution

k= (2mg(3.9cm+7.85cm) / (7.85cm^2)) = 32.36

f= k x

f= (32.36) (.039 - .0785) = 1.27822 N


correct answer is: 1.27 N

 

[PhanthomJay]

Homework Statement

When an 86.5-g piece of toast is inserted into a toaster, the toaster's ejection spring is compressed 7.85 cm. When the toaster ejects the toasted slice, the slice reaches a height of 3.9 cm above it's 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. Assume that throughout the ejection process the toast experiences a constant acceleration.

Homework Equations

Not sure. I'm confused about 'assumptions' with the question. Seems broken to me. I approached it using different ways; 1. finding the spring constant of the spring, plugging that into the force formula [f=kx] OR ignoring the fact that the spring is a spring, then comparing forces [setting them equal to each other] and solving for some variable. The first method seems to work, but has given twice the correct answer. I'm not looking for the answer, I'm looking for the methodology.

The Attempt at a Solution

k= (2mg(3.9cm+7.85cm) / (7.85cm^2)) = 32.36

f= k x

f= (32.36) (.039 - .0785) = 1.27822 N


correct answer is: 1.27 N

The spring exerts a force on the toast only when it is in contact with it over its compressed and uncompressed length, a distance of 0.0785 m. . The max force delivered by the spring is kx. Since it delivers no force when x = 0, the average force of the spring is half the max force.

 

[the1cyrus]

Thanks!

How would you find the time?

question: What is the time over which the ejection spring pushes on the toast. Assume that throughout the ejection process the toast experiences a constant acceleration.

The way I attempted solving this problem was to use acceleration = force divided by mass then use conservation of energy to determine the exit velocity by equating .5kx^2 = .5mv^2. Next, I found time by dividing velocity by acceleration to find time from equation v = axt; however, this is not providing the correct result which is 0.18s; I am coming up with .104 seconds. Any help would be great because I am pretty sure my physics is right.

 

[PhanthomJay]

the toast pops up vertically. so you must include its potential energy in your equation, which you neglected to do..

 

[asteriatic]

I would approach this problem by first identifying the known and unknown variables. The known variables are the mass of the toast (86.5 g), the compression distance of the spring (7.85 cm), and the height of the toast after ejection (3.9 cm). The unknown variables are the average force exerted by the spring on the toast and the time over which the spring pushes on the toast.

Next, I would use the known variables to solve for the spring constant (k) using the formula k = mg/x, where m is the mass of the toast and x is the compression distance of the spring. This gives a value of k = 32.36 N/m.

Using this spring constant, I would then use the formula F = kx to calculate the average force exerted by the spring on the toast. Plugging in the values for k (32.36 N/m) and x (0.0785 m), we get an average force of 1.27822 N. This is slightly higher than the correct answer of 1.27 N, which could be due to rounding errors.

To calculate the time over which the spring pushes on the toast, we can use the formula t = √(2x/a), where a is the acceleration of the toast and x is the height it travels. Since the problem states that the toast experiences a constant acceleration, we can assume that the acceleration is equal to the acceleration due to gravity (9.8 m/s^2). Plugging in the values for x (0.039 m) and a (9.8 m/s^2), we get a time of 0.2 seconds. This means that the ejection spring pushes on the toast for 0.2 seconds before releasing it.

In conclusion, the average force exerted by the ejection spring on the toast is 1.27822 N and the time over which the spring pushes on the toast is 0.2 seconds. The slight discrepancy in the force calculation could be due to rounding errors or uncertainties in the given values.