Calc Work Needed to Push Car up 11.0 deg Incline

In summary, the minimum work required to push a 1068 kg car 305 m up a 11.0 degree incline with an effective coefficient of friction of 0.26 is equal to the work required to compensate for gravity and friction forces. The normal force acting on the car is not equal to its weight, but rather the component of the weight that is perpendicular to the surface. This can be found by using the cosine of the incline angle. The theorem of variation of kinetic energy can be applied in this situation to find the net force along the direction of motion. However, since the initial and final velocities are not given, this method cannot be used. Therefore, Galileo's advice to apply the theorem of variation
  • #1
xelda
23
0
Whis is the minimum work needed to push a 1068 kg car 305 m up a 11.0 degree incline if the effective coefficient of friction is 0.26?

I have a hard time understanding why the normal force does not equal mg? How do I find the normal force? Or am I looking at this problem incorrectly?
 
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  • #2
Draw a picture of the car on the incline.
Gravity acts straight down vertically, but the normal force is perpendicular (normal) to the surface. Since the incline makes an angle with the vertical, the normal force acting on the car also makes an angle with the gravitational force.
Try to find the component of the gravitational force perpendicular to the surface (or parallel to the normal).
 
  • #3
Apply the theorem of variation of KE...It will be immediate.

Daniel.

P.S.Compute the forces correctly...Take Galileo's advice.He's really keen on inclines...:tongue2:
 
  • #4
dextercioby said:
Apply the theorem of variation of KE...It will be immediate.
What do you mean by this?
It is just a matter of net force along the direction of motion and distance also.
 
  • #5
The minimum work done will be the work one has to do is simply compensate the work done by gravity & frition force...Sure,in this case,bacause,the initial & final states are not specified (namely the veloities being given),one cannot use the theorem...

So my advice was not lucrative for this problem,sorry.

Daniel.
 
  • #6
I would assume the normal force is mg x sin theta, but I was told this was incorrect?
 
  • #7
xelda said:
I would assume the normal force is mg x sin theta, but I was told this was incorrect?
No man. If [tex]\theta[/tex] is the angle that the incline makes with the horizontal then you take [tex]cos\theta[/tex]! Draw the picture and see why.
 

1. How do you calculate the work needed to push a car up an 11.0 degree incline?

The work needed to push a car up an 11.0 degree incline can be calculated using the formula W = mgh cos θ, where W is the work (in joules), m is the mass of the car (in kilograms), g is the acceleration due to gravity (9.8 m/s^2), h is the height of the incline (in meters), and θ is the angle of the incline (in degrees).

2. What units are used to measure work in this calculation?

The units used to measure work in this calculation are joules (J). Joules are a unit of energy and are equal to one newton-meter (N*m).

3. Do you need to know the weight of the car to calculate the work needed?

Yes, the weight of the car (in newtons) is needed to calculate the work needed to push it up an incline. This can be calculated by multiplying the mass of the car (in kilograms) by the acceleration due to gravity (9.8 m/s^2).

4. Can this calculation be used for any incline angle?

Yes, this calculation can be used for any incline angle. However, it is important to note that the work needed will vary depending on the angle of the incline. The steeper the incline, the more work will be needed to push the car up.

5. How accurate is this calculation?

This calculation is accurate as long as the values used (mass, height, angle) are measured correctly. Any errors in measurement can affect the accuracy of the result. It is also important to consider other factors such as friction and air resistance, which may affect the actual amount of work needed to push the car up the incline.

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