Calculating Min Work for 800kg Car on 9.0° Incline

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I have been trying this problem multiple times but it still says I'm wrong:

What is the minimum work needed to push a 800 kg car 930 m up along a 9.0^\circ incline?

i'm using the formula:
W= Fdcos(theta)
F= mg

what am i doing wrong?
 
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Welcome to PF!

Hi cmed07! Welcome to PF! :smile:

(have a theta: θ and a degree: º :wink:)

First, what is the minimum possible value of F?
 
cmed07 said:
I have been trying this problem multiple times but it still says I'm wrong:

What is the minimum work needed to push a 800 kg car 930 m up along a 9.0^\circ incline?

i'm using the formula:
W= Fdcos(theta)
F= mg

what am i doing wrong?
There are a couple of ways to approach this. One way is to calculate the component of force (gravity) along (ie. parallel to) the [tex]9.0^\circ[/tex] inclined surface and multiply that force by the distance (930 m). The simpler way would be to determine the height increase over that 930 m and the resulting change in gravitational potential energy of the car. The work is equal to the change in gravitational potential energy.

AM
 


tiny-tim said:
Hi cmed07! Welcome to PF! :smile:

(have a theta: θ and a degree: º :wink:)

First, what is the minimum possible value of F?


That's all the information that was given to me... I know I'm supposed to find force by multiplying the mass and gravity...but i think the number I'm getting after i put it into the work formula is too big...
 
cmed07 said:
That's all the information that was given to me...

D'oh! :rolleyes:

On the information that was given to you, what is the minimum possible value of F? :smile: