Leg Presses. As part of your daily workout, you lie on your...

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do $80.0 \mathrm{~J}$ of work when you compress the springs $0.200 \mathrm{~m}$ from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform $0.200 \mathrm{~m}$ farther, and what maximum force must you apply?

Key Concepts

Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in its energy. In the context of spring mechanics, the work performed to compress a spring converts directly into its potential energy. This principle allows us to calculate the amount of work needed to achieve a certain compression, which in turn helps in determining the necessary force applied over a displacement.

Effective Spring Constants in Parallel

When springs are arranged in parallel, their individual spring constants add together to yield an effective spring constant for the system. This means that two identical springs in parallel act like a single spring with twice the spring constant of one. This concept is important when analyzing mechanical systems that use multiple springs in conjunction, as it affects both the required force and the energy stored for a specific displacement.

Hooke's Law

Hooke's law is the principle that the force exerted by a spring is directly proportional to the displacement from its rest position. This relationship is described by the equation F = kx, where k is the spring constant and x is the displacement. It serves as the foundation for understanding how springs resist compression or extension, which is critical when analyzing forces during exercises that involve spring-like mechanisms.

Spring Potential Energy

Spring potential energy refers to the energy stored in a deformed spring, which can be expressed as U = (1/2) kx². This concept explains how work done on a spring is converted into stored elastic energy. Understanding spring potential energy is essential for calculating the work required to compress or extend a spring by a given distance.

Transcript

00:01 Hello, my name is david.

00:02 In this video, we will cover the force and the work on a spring.

00:08 So for this problem, we're supposed to be doing leg presses on two horizontal springs.

00:14 So we're treating both springs at just one spring.

00:17 And when we compress the two springs, we apply a work of 80 joules.

00:27 And that's when we compress it, 0 .2 meters.

00:31 But since we're compressing it, our x is going to be negative.

00:33 0 .2 meters.

00:37 So equation 8 gives you the force to stretch or to compress the spring and that's equation 6 .8.

00:49 So we must know the spring constant first.

00:53 So to find that we're going to apply the equation 6 .9 which gives the work on spring when you stretch it from its equilibrium position, which is 1 half k x square.

01:08 So we saw for k, we must multiply both sides by 2 and divide by x square so we have k equals 2 w divided by x square so we could plug in our values so 2 times 80 joules divided by negative 0 .2 meters square so we see that our spring constant it's 4 000 newtons per meter okay so now we could plug this into equation 6 .8.

02:07 So the force equals the spring constant times the distance stretch or compressed from its original position, i mean in equilibrium position.

02:18 So it's 4 ,000 neutrons per meter times negative 0 .2 meters.

02:28 So the force required to compress the spring, it's negative 800 newtons.

02:36 But for this part, they're asking us for the magnitude of the force.

02:41 So the magnitude through this force is just 800 newtons.

03:04 Now for part b, we want to know how much additional work is required to compress it an extra 0 .2 meters.

03:17 So for this one, we're going to apply equation 6 .10, which gives you the work on a spring when compressing it or stretching it from an initial position, which is not the equilibrium position...

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