Eric
model lt1024 cub cadet 24hp 450hours very good condition
Answer
Beats me. Make believe you are trying to buy one, check on Craigslist and your local Want Advertiser, see what others are asking, and price yours accordingly.
Beats me. Make believe you are trying to buy one, check on Craigslist and your local Want Advertiser, see what others are asking, and price yours accordingly.
A 120 kg lawn tractor goes up a 21 degree incline of 12.0m in 2.5s. What power is shown by the tractor?
A
Why do you use sine in this question to figure out the vertical component of the force? Why do you not use the 12.0 m that's given because isn't that the displacement?
or why don't you use the horizontal component of the force??
Answer
Power = Work/Time.
The definition of Work = (size of Force) x (size of Displacement) x (cos Î)
Where Force and Displacement are vector quantities.
In 2-D physics the vector Force and vector Displacement are in the same plane so that there is a single angle = Î existing between them.
When Π= 90° or 270° the Work is zero.
When 0 ⥠Π< 90° or 270° > Π⤠360° the Work is > 0
When 90° < Π⤠180° or 180 ⤠Π< 270° the Work is < 0
So to answer UR question:
In order to calculate Work U can calculate it one of three(3) different ways:
In two of these ways, the angle Πis forced equal to 0°. The COMPONENT of either Force or Displacement is taken in the other's direction forcing Πin the Work formula = 0.
These are the TWO component ways to calculate the work it takes tractor to move up incline:
1) take either the Force (component) parallel to the Displacement or 2) the Displacement (component) parallel to the Force.
THIS is what is happening when the Displacement which is parallel to the incline and the Force which is the weight (mg) of the tractor in the vertical in direction.
1) (mg) (sin 21°) x 12.0 = work of tractor up incline
2) (mg) x 12.0 (sin 21°) = work of tractor up incline
The third way to calculate the work is just to use the Work formula as is:
3) (mg) x (12.0) x cos (90°-21°) = (mg)(12.0)(cos 69°) = work of tractor up incline
comment: this 3rd way may be more difficult in that the value of Î must be defined
Power = Work/Time.
The definition of Work = (size of Force) x (size of Displacement) x (cos Î)
Where Force and Displacement are vector quantities.
In 2-D physics the vector Force and vector Displacement are in the same plane so that there is a single angle = Î existing between them.
When Π= 90° or 270° the Work is zero.
When 0 ⥠Π< 90° or 270° > Π⤠360° the Work is > 0
When 90° < Π⤠180° or 180 ⤠Π< 270° the Work is < 0
So to answer UR question:
In order to calculate Work U can calculate it one of three(3) different ways:
In two of these ways, the angle Πis forced equal to 0°. The COMPONENT of either Force or Displacement is taken in the other's direction forcing Πin the Work formula = 0.
These are the TWO component ways to calculate the work it takes tractor to move up incline:
1) take either the Force (component) parallel to the Displacement or 2) the Displacement (component) parallel to the Force.
THIS is what is happening when the Displacement which is parallel to the incline and the Force which is the weight (mg) of the tractor in the vertical in direction.
1) (mg) (sin 21°) x 12.0 = work of tractor up incline
2) (mg) x 12.0 (sin 21°) = work of tractor up incline
The third way to calculate the work is just to use the Work formula as is:
3) (mg) x (12.0) x cos (90°-21°) = (mg)(12.0)(cos 69°) = work of tractor up incline
comment: this 3rd way may be more difficult in that the value of Î must be defined
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