# Work, Energy, and Force

1. Nov 10, 2004

### M1zzSandra

hey!
At school, im currently studying work and energy. It involves forces too which i had trouble with. I don't understand the concept of work and energy. The teacher, he gives us problems with work and energy and expects us to know what is going on. Some problems that have a loop, like a roller coaster, and i would have to find the speed and stuff. Arghh, the whole work, energy, and force is so complicated. Is there some way to make it easier for me to understand? ><

2. Nov 10, 2004

### Math Is Hard

Staff Emeritus
What do you find the most confusing? You know that energy can be described as the ability to do work. You know that work is force times distance. You know that force is mass times acceleration. The roller coaster problem sounds like you'll be doing some measurements of kinetic and potential energy. Why don't you post one of your problems here and the good folks of this board will be eager to help!

3. Nov 11, 2004

### M1zzSandra

what kind of problems and when will you know that you have to use the equation W = KE + PE?? and what kind of problems will you know that you have to use the equation W = (KEf - KEo) + (PEf - PEo)???

4. Nov 11, 2004

### rcgldr

Ok, some simple concepts.

I'm assuming you understand force. You can have a force with no movement, you can stand on a scale to meaure weight.

Work is defined as force times distance, for example, lb ft.

Power is defined as a rate of work per unit time, for example

$${lb\ ft}/sec$$

Power can be interpreted as a force times speed, for example, horse power can be calculated as follows:
$$1\ hp = 550\ lb\ ft\ /\ sec$$

$$Power_{hp} = Force_{lb}\ x\ Speed_{mph}\ / \ 375$$

Energy can be thought of as the net work or work potential of an object. For example, an object at rest (with respect to some frame of reference) has no kinetic energy (with respect to that same frame of reference). Apply work to that object to accelerate it, and it's kinetic energy will be equal to the amount of work applied to the object (assuming no losses here). For a specific example, assume a 1 slug (= 32.174 lb) mass is acclerated with 1 lb of force for a distance of 32 feet. You end up with 32 lb ft of kinetic energy.

The object is being accelerated at 1 ft / second. Distance = 1/2 a t^2, so doing the math, the time is 8 seconds. Velocity will be 8 ft / sec. Kinetic energy can be calculated as 1/2 m v^2, in this case 1/2 x 1 x 8^2 = 32 lb ft, which corresponds with the previous calculation.

Now do the same with a 1 lb = 1/32.174 slug mass. Object accelerates at 32.174 ft/sec^2, or 1 "g". Still 32 lb ft of work done, time is 1.41 seconds, and speed ends up 45.38 ft / sec. Recalculating, 1/2 x 1/32.174 x 45.38^2 = 32.

To understand potential energy, assume the 1 lb object is 32 feet above the ground (which is at sea level, so gravitational accleration is 32.174 ft /sec^2). The potential energy is considered to be 32 lb ft, as this is what the kinetic energy would be by the time the dropped object hits the ground. If the altitude is small enough that changes in gravitational strength can be ignored, then the potential energy = mass times altitude. From the time the object is dropped until it hits the ground, the sum of potential and kinetic energy will be the same, always 32 lb ft in my example.

Even if we take into account that gravitational strength decreases with distance, the sum of potential and kinetic energy will be a constant, as long as no other forces (ignoring aerodynamic drag here) are involved. Because gravity decreases with the square of the distance (from the center of masses), there's a maximum potential energy. In the case of the earth, if an object at rest (with respect to earth) were an infinite distance away from earth, and dropped, it's maximum velocity would be 36745 ft/sec at the center of the earth if it could fall through a tunnnel to reach the center of earth, a bit less if it impacted on the surface.

Last edited: Nov 11, 2004