2007-01-07 11:06:58 · 10 answers · asked by fahed 1 in Science & Mathematics ➔ Physics
there will have been no work done, but there still would be a loss of energy.
no work is done (in a physics sense) because work=force*distance. if we say this body traveled 4 m out, then 4m back (4m in the positive and 4m in the negative direction, thus 0m total displacement), the total distance traveled is 0. thus the work done would be 0.
2007-01-07 11:10:40 · answer #1 · answered by Aubrey D 2 · 0⤊ 1⤋
Work is force multiplied by distance
Force is mass multiplied by acceleration
Acceleration is velocity divided by time
Velocity is a VECTOR... the direction of the velocity is integral to the quantity being measured, thus by changing direction you change the velocity which means there is an acceleration which means that force must be applied and therefore work is done.
You need to apply one force to get from a to b, and *another* to slow down when you arrive at b, then you need yet another force to get from b to a, therefore the sum is not zero, it is in fact double the work done for a single trip (assuming you stop when you get back to point a).
Shame it's not zero... otherwise when I drive to work in the morning I would be emptying the fuel tank, but when I drive back home the fuel tank would be filling up again!
2007-01-07 23:24:08 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Work = Force x Distance
For example, if the distance from point a to b = 2 meters
from a to b = 2 meters
from b to a = -2 meters
If you add the total distance from a to b and from b to a, it equals zero. Then, of course, when you multiply the force by zero you still have zero.
2007-01-07 11:16:34 · answer #3 · answered by WxEtte 5 · 0⤊ 1⤋
They are back to where they started. I would say they have traveled 2 times the distance from a to b. They have not gone anywhere but have wasted energy. They are still at the same point at the end.
2007-01-07 11:09:06 · answer #4 · answered by CrazyChic86 3 · 1⤊ 0⤋
It depends on the environment. If you are doing work in a conservative field, then the net work performed for the round-trip is zero; for example, taking a round trip in a perfect gravitational field. On the other hand, if the field is not conservative, or there is a dissipative force such as friction that you work against the entire round trip, then the net work will be non/zero for the round trip.
2007-01-07 12:12:12 · answer #5 · answered by vejjev 2 · 3⤊ 1⤋
No, the work will be a to b, then b to a. Or (a to b)^2....
Nothing moves with out work / energy input.
If you figure that it does, say nothing to nobody, and come tell me in secret, and we shall make millions....
2007-01-07 11:11:34 · answer #6 · answered by Anonymous · 1⤊ 1⤋
let x be distance
Displacement = zero
Work to b= force x distance = x
Work to a = force x distance = -x
Therefore OVERALL work is 0 but work was done when breaking it up.
2007-01-07 12:06:22 · answer #7 · answered by life_aint_a_game_10 2 · 0⤊ 1⤋
The displacement would be 0, I imagine work would be doubled.
2007-01-07 11:14:43 · answer #8 · answered by itsmalkara 2 · 1⤊ 1⤋
no as you expended energy to move back and forth.
2007-01-07 11:11:36 · answer #9 · answered by dude_port 3 · 1⤊ 1⤋
work was done , though there was no net movement
2007-01-07 11:20:45 · answer #10 · answered by tyab26 1 · 2⤊ 1⤋