Power/work questions (with inclines)

  • Thread starter dnt
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In summary, when calculating work done by a force on an object, the cosine is used to get the component of the force parallel to the displacement. In the given question, the force pushing the tractor up the incline has to cancel the weight component, which is given by the sine of the inclination. Similarly, in the second question, the work done is calculated by getting the component of the force parallel to the displacement, which again involves using the cosine. The reason for this is to get the accurate component of the force that is doing work as the object is displaced.
  • #1
dnt
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the question states at 120 kg tractor goes up a 21 degree incline of 12.0 m in 2.5 seconds. what is the power?

ok i know power is W/T which is basically Fd/T

my question is how do you use the incline. work is Fdcos0 but i just don't see why its cosine in this question. the work is going UP so shouldn't you want to use the vertical component? meaning use sine insteand of cosine? doesn't make sense to me.



a similar question is a 185 fridge goes up 10.0 m at an angle of 11 degrees, what is the work done?

again, its going up so should we use cosine or sine? i think it should be sine but the equation says use cosine.

dont quite get it. please help. thanks.
 
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  • #2
The work under discussion here is that done by the force pushing the object up the incline. This force has to cancel the weight component of the object, parallel to the incline, out. Is this component given by the sine or the cosine of the inclination?
 
  • #3
i believe that's the sine. is that correct?
 
  • #4
Correct. So the you want to calculate the work done by a force equal to [itex]W\sin(\theta)[/itex], but in the opposite direction, that pushes the object up the incline. The cosine in the work equation is applicable when the angle [itex]\theta[/itex] is given as the angle between the force and the displacement vectors. The reason why there is a [itex]\cos(\theta)[/itex] in the work equation is to get the component of the force that is parallel to the displacement. It is only this component that does work as the object is displaced. The other component, perpendicular to the displacement, does no work as the object is displaced.
 
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  • #5
Just use potential energy. But I'll show you why you use cosine anyway.
[tex]W = \int_{s_i}^{s_f} F_s ds = \int_{s_i}^{s_f} F cos(\theta)ds[/tex]
F and [tex]\theta[/tex] are constant and can be moved out of the integral. F_s is the component of F(vector) in the direction of motion, and F makes an angle [tex]\theta[/tex] with respect to a displacement [tex]\Delta r[/tex], so the component of the force vector is the direction of motion is [tex]F_s = Fcos\theta[/tex].
[tex]W = Fcos(\theta)\int_{s_i}^{s_f}ds = F cos\theta (s_f - s_i) = F cos\theta (\Delta s) = F(\Delta r) cos\theta[/tex]
 
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  • #6
andrevdh said:
Correct. So the you want to calculate the work done by a force equal to [itex]W\sin(\theta)[/itex], but in the opposite direction, that pushes the object up the incline. The cosine in the work equation is applicable when the angle [itex]\theta[/itex] is given as the angle between the force and the displacement vectors. The reason why there is a [itex]\cos(\theta)[/itex] in the work equation is to get the component of the force that is parallel to the displacement. It is only this component that does work as the object is displaced. The other component, perpendicular to the displacement, does no work as the object is displaced.

so isn't that a little misleading? why do they include cosine in the work equation when sometimes you use sine?
 
  • #7
When we want to calculate the work done by a force applied to an object we need to get the component of the force that is in line with the displacement - therefore the [itex]\cos(\theta)[/itex] in the work equation. Where [itex]\theta[/itex] is the angle between the direction of the force and displacement vectors. In this case we've got the force parallel to the incline, so there is no need for the cosine. The sine comes from gettting the component of the weight parallel to the incline, which involves the angle of the incline, which is a completely different angle than the one in the work equation.
 

1. What is the relationship between power and work on an inclined plane?

The relationship between power and work on an inclined plane is that power is the rate at which work is done or energy is transferred, and work is the energy expended to move an object against a force. On an inclined plane, the work done is equal to the force applied along the plane multiplied by the distance moved along the plane, while power is equal to the work done divided by the time it takes to do the work.

2. How do you calculate the power required to move an object up an inclined plane?

The power required to move an object up an inclined plane can be calculated by multiplying the force needed to move the object along the plane by the velocity at which the object is moving, or by dividing the work done by the time it takes to do the work.

3. What is the effect of increasing the angle of inclination on the power required to move an object up the inclined plane?

Increasing the angle of inclination on an inclined plane will increase the power required to move an object up the plane. This is because as the angle increases, the force needed to move the object along the plane also increases, resulting in a greater amount of work being done and therefore a higher power requirement.

4. Can power be negative when calculating work on an inclined plane?

Yes, power can be negative when calculating work on an inclined plane. This occurs when the force applied to move the object is in the opposite direction of the movement, causing the work done to be negative. In this case, the power will also be negative, indicating that energy is being taken away from the object rather than being transferred to it.

5. How does the weight of an object affect the power needed to move it up an inclined plane?

The weight of an object does not directly affect the power needed to move it up an inclined plane. The power required is dependent on the force applied to move the object along the plane and the speed at which it is moved. However, the weight of the object does play a role in determining the force needed to move it along the plane, as a heavier object will require a greater force to overcome its weight and move it up the plane.

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