How much work does gravity do on the car

[Brittykitty]

A car of mass 1500 kg rolls down a hill for a distance of 150 m.
If the road makes an angle of 15º with the horizontal, how much work is done on the car by the force of gravity?

would I use the formula

W=fdcosº
w=(f)(150m)(15º)


For force would I convert 1500kg into Newtons and plug that into the formula?


Thanks :)

 

[Mintea]

I believe you're right about the force, but are you sure you're supposed to take the cosine of 15º? Since Work also equals the change in Potential Energy, it seems like you'd only want the vertical distane the car travels.

(note: not at all to scale)
`|\
`|`\
y|``\ 150m
`|```\
`|___(\<- 15º
```x

150m * cos 15 would give you x; multiplying 150m by either cos 75 or sin 15 would give you the proper value, depending on which way you like to do it.

 

[rock.freak667]

Firstly, you need to get the component of the force in the direction of motion. It is not just as easy as Fdcosθ.

 

[sdcraigcooper]

In this case, I think that you would need, at some point, to take the mass times g ( 9.8 m/s^2 )
I base this assumption on this fact: the unit of force is Newton, which is the same as kg - m/s^2

SO... W=(f)(d)cos(theta) as you said above. plug in the variables as you have said, expect use the weight (mass times g) for F...

Also, don't forget: work can also mean, more simply, change in energy. energy conservation usually provides an easier solution.

 

[Brittykitty]

In this case, I think that you would need, at some point, to take the mass times g ( 9.8 m/s^2 )
I base this assumption on this fact: the unit of force is Newton, which is the same as kg - m/s^2

SO... W=(f)(d)cos(theta) as you said above. plug in the variables as you have said, expect use the weight (mass times g) for F...

Also, don't forget: work can also mean, more simply, change in energy. energy conservation usually provides an easier solution.

Hello :)

Thank you! So the correct equation would be:
w=(14715)(150m)(cos15)?

 

[rock.freak667]

Hello :)

Thank you! So the correct equation would be:
w=(14715)(150m)(cos15)?

Please draw over your free-body diagram, if you do, you will see why it is not cosine.

 

[Brittykitty]

Would it be sin?

 

[sdcraigcooper]

The force moving the car down would the the sine component of the force, or mgsin(theta)

 

[rock.freak667]

Would it be sin?

Yes it is sine but it is best to understand why it is sine and not cosine.

 

[sdcraigcooper]

Yes it is sine but it is best to understand why it is sine and not cosine.

I agree. If you decompose the force of weight on the car, which would point striaght down, it breaks up into 2 parts: one parallel with the surface normal (or normal force) and one perpindicular to the surface normal. The one that is actually pulling the car down the slope is the force that is perpindicular to the surface normal. Here is a picture... I hope it helps...

http://www.facebook.com/photo.php?pid=268554&id=100000066426322

 

[Brittykitty]

The picture helps a ton!
W=fdcos
w=(14715)(150m)(sin15)
w=571278

does that seem a bit high?

 

[rock.freak667]

The picture helps a ton!
W=fdcos
w=(14715)(150m)(sin15)
w=571278

does that seem a bit high?

Yes that is correct. But remember that 150 m is a long distance so the work done will be high.

 

[Brittykitty]

That is true, thank you so much Rock Freak :)