- #1

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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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- Thread starter cmed07
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- #1

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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?

- #2

tiny-tim

Science Advisor

Homework Helper

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Hi cmed07! Welcome to PF!

(have a theta: θ and a degree: º )

First, what is the minimum possible value of F?

- #3

Andrew Mason

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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.

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?

AM

- #4

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- 0

Hi cmed07! Welcome to PF!

(have a theta: θ and a degree: º )

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...

- #5

tiny-tim

Science Advisor

Homework Helper

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That's all the information that was given to me...

D'oh!

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