How much work can be done given force, mass, angle & displacement?

[fixedglare]

Homework Statement

a person uses a cord to pull a boat of 1000 kg along 50 meters. The cord makes a 45° angle. If the applied work to the boat is 40 N, how much work is done?

Given:
mass = 1000 kg
displacement = 50 meters
the angle = 45°
applied work = 40 N

Homework Equations

Work = Fd
Weight = mg
Work = Fd cos θ

The Attempt at a Solution

First I converted the mass to weight:

1000 kg/ 9.81 m/s = 101.9 N

Then I added that to the other force so, 101.9 N + 40.0 N = 141.9 N. *I don't know if I should add or subtract*

Then I used the W = Fd cos θ equation:

W = (141.9 N * 50 m)(cos 45°) =

5016.9 JIs this correct?

 

[jackarms]

You can't just add the weight and the applied force together. They're vectors, so they have to be handled separately, or by components. Calculate work done by each force separately, and by the two components of the applied force, and then add those together into the net work.

 

[fixedglare]

You can't just add the weight and the applied force together. They're vectors, so they have to be handled separately, or by components. Calculate work done by each force separately, and by the two components of the applied force, and then add those together into the net work.

Is it,

(101.9 N (cos 45)) + (40 N (cos 45)) = 100.4 N

100.4 N * 50 m = 5019.2 J

Yes?

 

[jackarms]

No, you have it a bit mixed up. Just think about the problem for a second. The displacement is straight to the right. The weight points straight down. The applied force points up and to the right, at a 45 degree angle. Calculate work done by weight as:$$W_{weight} = mg \cdot d \cdot cos\alpha$$ and $$W_{app} = F_{app} \cdot d \cdot cos\beta$$ where \alpha is the angle between the weight force and the displacement, and \beta is the angle between the applied force and the displacement.

 

[fixedglare]

No, you have it a bit mixed up. Just think about the problem for a second. The displacement is straight to the right. The weight points straight down. The applied force points up and to the right, at a 45 degree angle. Calculate work done by weight as:$$W_{weight} = mg \cdot d \cdot cos\alpha$$ and $$W_{app} = F_{app} \cdot d \cdot cos\beta$$ where \alpha is the angle between the weight force and the displacement, and \beta is the angle between the applied force and the displacement.

So;

Wweight = 1000 kg * 50 m * cos 45? 35355.3

+

Wapp = 40 N * 50 m * cos 45? 1414.2

1414.2 + 35355.2 = 36769.4 J ?

 

[BvU]

Funny calculation. is the boat really being pulled up? i.e. is the displacement in the vertical direction also 50 m? Or are we pulling a boat along a canal over a towpath (explains the 45 degrees -- keeps the feet dry) and is the water level at the start the same as at the destination ?

 

[fixedglare]

Funny calculation. is the boat really being pulled up? i.e. is the displacement in the vertical direction also 50 m? Or are we pulling a boat along a canal over a towpath (explains the 45 degrees -- keeps the feet dry) and is the water level at the start the same as at the destination ?

the boat is being pulled along the canal by the person towards the board walk. The problem in my book does not say whether the water level is at the same as at the destination so I presume it is.

 

[BvU]

And: what's this "applied work to the boat is 40 N" ?? Work is Newtons times meters. We interpret the 40 N as a force of 40 N, but is this correct ? 40 N is like pulling with one finger...

 

[BvU]

the boat is being pulled along the canal by the person towards the board walk. The problem in my book does not say whether the water level is at the same as at the destination so I presume it is.

So: no work done there. 35355.2 J saved.

By the way: if your given data are 1 or 2 digits, don't present the final result with 6 digit precision. 35 kJ looks a lot better (but we just erased that...).

 

[fixedglare]

Gah, I'm confused. I'm sorry if it doesn't make sense I'm translating it to English.

I'll restate it, this time more carefully.

a person uses a cord to pull a boat that has a mass of 100 kg along 50 m by 50 m. along the board walk. The cord makes a 45 degrees angle. If the applied force to the boat is 40 N, how much work is realized?

What I did (with the help of jackarms) was to calculate each force individually using the equation W = (F) (d) (cos θ) and then add them both for the result.

 

[fixedglare]

So: no work done there. 35355.2 J saved.

By the way: if your given data are 1 or 2 digits, don't present the final result with 6 digit precision. 35 kJ looks a lot better (but we just erased that...).

What do you mean saved? How do I solve it?

IDK. The other one made sense to me.

 

[BvU]

Yes, so the boat doesn't move up or down: Wweight path is 0 m. No Wweight. Only Wapp. Checked the number, same result. One finger can do the work ;-)

 

[BvU]

IDK meaning ?

 

[gneill]

So a picture of the problem would look something like this?

Presumably the only relevant motion is horizontal (50 m). The weight of the boat is supported by buoyancy, and no work is needed to float the boat What portion of the 40 N force is acting in the direction of motion?

 

[fixedglare]

IDK meaning ?

Then how do I do it?

 

[gneill]

IDK meaning ?

Text speak: IDK = I Don't Know. "Text speak" is not allowed here. fixedglare, please use proper English and grammar for posts in the forums.

 

[BvU]

Dumb me no tekst speak. I also drew a picture. No bird pulling the boat, just a person -- with feet dry. Cheaper boat - same result.

 

[fixedglare]

Text speak: IDK = I Don't Know. "Text speak" is not allowed here. fixedglare, please use proper English and grammar for posts in the forums.

I'm sorry, I'll be more careful.

 

[gneill]

Dumb me no tekst speak. I also drew a picture. No bird pulling the boat, just a person -- with feet dry. Cheaper boat - same result.