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Review of mechanical work

by jschmidt
Tags: mechanical, review, work
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jschmidt
#1
Aug18-11, 03:32 PM
P: 22
Hi,

It's been a few years since my physics classes, and I want to do a quick double check to make sure I correctly remember the fundamentals of mechanical work.

1) At the most basic level, work is a force applied to an object, across a distance (that is, work cannot be calculated where there is no motion), yes? In the simplest case, where an object is experiencing linear motion, and the direction of force is either in the exact same direction, or the exact opposite direction, this corresponds to the formula:

Work = Force * distance, where work, force, and distance are simple scalar values (not vectors).

If you need to get fancier, you can use vectors (either 2D or 3D), with corresponding formulas which basically are the same thing, but take into account the direction of the force, versus the direction of the motion (only the fractional component of the force which is in the same or opposite direction as the motion is multiplied by the scalar of the distance, to produce work; this can be done using either trigonometry with angels, or with vector products [dot-product or cross-product, don't remember right now, but I can look it up later])?

2) In a system where multiple forces are acting upon an object, we can calculate the individual work of each force, and also a "net work", which is the work accomplished by the "net force", which is the sum of all forces.


So, for example, a person in a small boat with an outboard motor, going upstream in a river (assuming a 'simple' current model where there's no eddies, etc, but all of the force of the current is in the same direction, downstream). The motor produces a force upon the boat, in the upstream direction. The current produces a force on the boat in the downstream direction. For the boat to move upstream, the force of the motor must be greater than the force of the stream.

If the boat moves 10 meters upstream, we can say that the motor did 10F_motor Joules of work (in this case, not taking the net force into account)? Is that a valid statement? Or does "work" only apply to F_net * distance? I think we can also say that the stream did 10F_stream joules of negative work on the boat? Together, the Net work = 10 (F_motor - F_stream)?
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IttyBittyBit
#2
Aug18-11, 07:27 PM
P: 159
Yes that is correct as long as the boat is not accelerating (i.e. moving at a steady velocity). In addition, in that case you will have F_motor = -F_water (i.e. the force of the motor is just enough to overcome the drag of the water at that speed), so there will actually be no net work. Which is just another way of saying that no outside energy is added to the system, only energy from the motor.
jschmidt
#3
Aug18-11, 08:08 PM
P: 22
Quote Quote by IttyBittyBit View Post
Yes that is correct as long as the boat is not accelerating (i.e. moving at a steady velocity). In addition, in that case you will have F_motor = -F_water (i.e. the force of the motor is just enough to overcome the drag of the water at that speed), so there will actually be no net work. Which is just another way of saying that no outside energy is added to the system, only energy from the motor.
Why can't the boat be accellerating? I never said it was moving at a constant speed. It's my understanding - is this wrong - that in general, even if accelleration is happening, the work is still just the Force (or net Force, depending on what you're trying to calculate) times the distance travelled, is it not?

I mean, F_motor could even be < F_water - let's assume a situation where at the beginning of our measurements, the boat is travelling at some significant speed. Since the force of the motor is less than the force of the water, the boat will slow down, but isn't it true that the motor is *still* doing work, and the work the motor does is F_motor * distance? That the work the water does is -F_water * distance (or if you prefer, F_water * -distance), and that the Net work done by the entire systems is (F_motor - F_water) * distance, which since F_water > F_motor, will yield a negative value for the work (which negative value has the physical meaning that the boat was decelerated over the span traversed)?

IIRC, in an idealized system where there is no friction, and no other opposing forces, work *always* results in acceleration or deceleration? In the real world, whether it accelerates, decelerates, or the speed remains constant is a function of the *net* force - if net force is positive, there should be accelleration, if it's zero, speed remains constant, negative means deceleration (or acceleration in the opposite direction, if speed has reached 0)?

IttyBittyBit
#4
Aug18-11, 09:31 PM
P: 159
Review of mechanical work

Yes that is true, I was just making the observation that if the boat is accelerating the system is not in equilibrium - In other words, the forces do not balance, and thus the work done by the motor is greater that the water drag times the distance traveled. You need to take into account the inertial opposing force by the boat itself.


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