Works with coefficient of friction

[Echoeric666]

What is the minimum work needed to push a 1000 kg car up 45.0 meters up a 12.5 [tex]\circ[/tex] degree incline?
a. Ignore Friction
b. Assume the effective coefficient of friction is 0.30.


Work: W = (F*d)cos[tex]\theta[/tex]
Coefficient of friction: W = [tex]\mu[/tex]N(d)


What I did: W = (m*g)(d)cos[tex]\theta[/tex]. (1000kg)(9.8m / s^2)(45.0m)cos(102.5)
= -95449.9 J = 9.5 X 10^4 J work done to push the car up.

Then, the work friction is doing to car:
W = 0.30(1000kg*sin(12.5)*9.8 m/s^2)(45m)? This is where I get stuck because I know work done by friction on ramp is [tex]\mu[/tex]*normal force*displacement...

 

[rock.freak667]

What I did: W = (m*g)(d)cos[tex]\theta[/tex]. (1000kg)(9.8m / s^2)(45.0m)cos(102.5)
= -95449.9 J = 9.5 X 10^4 J work done to push the car up.

The vertical distance the car must move is not 45cosθ. Recheck this component. When you get the vertical height 'h', it just becomes W=mgh

Then, the work friction is doing to car:
W = 0.30(1000kg*sin(12.5)*9.8 m/s^2)(45m)? This is where I get stuck because I know work done by friction on ramp is [tex]\mu[/tex]*normal force*displacement...

What is the normal force at the angle θ ?

 

[Echoeric666]

Sorry, I meant 45 meters up as displacement. So I used that displacement in W = (F*d)cos[tex]\theta[/tex] as shown in post.

Also...you're saying if I find the height, I can use mgh to find works? (That potential energy is equal to work)?

 

[rock.freak667]

Sorry, I meant 45 meters up as displacement. So I used that displacement in W = (F*d)cos[tex]\theta[/tex] as shown in post.

Also...you're saying if I find the height, I can use mgh to find works? (That potential energy is equal to work)?

Yes since at the bottom there is only PE and at the top there will be only PE (since the mass will not be moving then), hence the change in PE will be the same as the work done. (This is because work done in a gravitational field in independent of path)

 

[rdl]

To calculate the minimum work needed to push the car up the incline, we need to consider both the work done against gravity and the work done against friction. The answer will differ depending on whether we ignore friction (a) or assume a coefficient of friction (b). Let's look at both scenarios:

a. Ignore Friction:
If we ignore friction, the minimum work needed to push the car up the incline is simply the work done against gravity. Using the equation W = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the incline, we get:
W = (1000 kg)(9.8 m/s^2)(45.0 m)cos(12.5°) = 430,974 J

b. Assume a coefficient of friction:
If we assume a coefficient of friction of 0.30, we need to take into account the work done against friction in addition to the work done against gravity. The equation for this would be:
W = (m*g)(d)cos\theta + \muN(d)
We already have the first part of the equation from part (a), so we just need to calculate the work done by friction. Using the equation W = \muN(d), where \mu is the coefficient of friction, N is the normal force (equal to the weight of the car in this case), and d is the displacement along the incline, we get:
W = 0.30(1000 kg)(9.8 m/s^2)(45.0 m)sin(12.5°) = 12,411 J
Therefore, the minimum work needed to push the car up the incline with a coefficient of friction of 0.30 is:
W = 430,974 J + 12,411 J = 443,385 J