Скачать или смотреть Force and Work (lifting problems), Single Variable Calculus

Let's look at the work done when lifting different objects (ropes, bags of sand). We see that work is the integral of F(x) from x=a to x=b, where F(x) represents the force on an object as it is displaced from x=a to x=b.

A force is essentially an influence that causes an object to undergo a displacement, akin to a push or a pull that moves an object. An everyday manifestation of force is the gravitational force that acts downwards on an object. The unit of force, internationally recognized, is the Newton N, defined as the force required to accelerate a one-kilogram mass at a rate of one meter per second squared (kg m/s^2).

Work, on the other hand, is the energy transfer that occurs when a force causes an object to displace. To have work, you need the force to move the object, and no movement means no work is done. Work is measured in Joules J, a unit denoted as a Newton meter (Nm), signifying the energy transferred when a one-Newton force displaces an object by one meter.

(While I usually refer to international units, it's worth mentioning that in the United States, force is measured in pounds, and work in foot-pounds. A pound reflects the force of gravity on an object's mass.)

00:00 Intro
02:59 Constant Force Examples
06:58 Top of the Building Version 1
19:07 Top of the Building Version 2
25:02 Leaking Sack of Sand Example

Let's consider a simple example that illustrates the relationship between force and work without involving calculus. Suppose you exert a force of 15 Newtons on a rock, causing it to displace sideways by three meters. The work done by this force is the product of the force and the displacement, which in this case amounts to 45 Joules.

Further examples will explore scenarios of constant force, such as lifting an object against gravity. Utilizing Newton's second law, we'll calculate the force exerted on an object as the product of its mass and the acceleration due to gravity. For instance, lifting a three-kilogram object by 20 meters would require a force equal to its mass multiplied by the gravitational constant (9.8 meters per second squared), resulting in a work output of 588 Joules.

We'll also address a lifting problem with a twist: imagine a rope hanging from a tall building. As the rope is wound up, the mass of the rope changes depending on how much of it is left hanging. This creates a variable force problem where different segments of the rope travel different distances. To solve this, we could set up a Riemann sum that, through the limit process, becomes an integral. The work done is then the integral of the force over the displacement. This approach can also be applied to a leaking bag of sand being lifted, where the force changes as sand leaks out.

When dealing with variable force problems, the general principle is to integrate the force function over the domain of displacement. This expands the basic definition of work beyond mere force times displacement, accommodating scenarios where force varies over the distance.

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