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 $$\circ$$ degree incline?
a. Ignore Friction
b. Assume the effective coefficient of friction is 0.30.


Work: W = (F*d)cos$$\theta$$
Coefficient of friction: W = $$\mu$$N(d)


What I did: W = (m*g)(d)cos$$\theta$$. (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 $$\mu$$*normal force*displacement...

 

[rock.freak667]

What I did: W = (m*g)(d)cos$$\theta$$. (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 $$\mu$$*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$$\theta$$ 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$$\theta$$ 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)