# Work done on an Object - Pulling a wagon while lifting up at an angle

• jigsaw21
In summary, Juri is pulling her wagon to the wagon repair shop with a force of 200 Newtons at an angle of 35 degrees to the horizontal. The work done on the wagon (in Joules) is 819.2 when using the official solution, but 573.6 when using sin(theta).
jigsaw21
Homework Statement
This question is asking to find the work done on an object being pulled at an angle.
Relevant Equations
W = F*d*cos theta
Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 Newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop?

...

Attempted Solution: I simply plugged in the values given into the formula above for Work which is F*d*cos (theta) due to the question saying that she was pulling at an angle of 35 degrees to the horizontal. This gave me an answer of 819.2 kJ. However, in the solution it is saying the appropriate formula should be W = F*d*sin (theta) instead of cos (theta), and thus the solution is 573.6 kJ.

But my question is why is sin(theta) used instead of cos(theta) for the formula . I thought that if the direction the wagon is being pulled needs to be parallel to the horizontal component of the Force, which should be cos (theta)?

I'm just curious if anyone could help me see why W = F*d*sin(theta) is the appropriate formula for this example and not W = F*d*cos(theta).

I appreciate any response or feedback.

Last edited by a moderator:
jigsaw21 said:
Homework Statement:: This question is asking to find the work done on an object being pulled at an angle.
Relevant Equations:: W = F*d*cos theta

Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahead of her--five kilometers--and she's pulling with a force of 200 Newtons. If she's pulling at an angle of 35 degrees to the horizontal, what work will be exerted on the wagon to get to the repair shop?

...

Attempted Solution: I simply plugged in the values given into the formula above for Work which is F*d*cos (theta) due to the question saying that she was pulling at an angle of 35 degrees to the horizontal. This gave me an answer of 819.2 kJ. However, in the solution it is saying the appropriate formula should be W = F*d*sin (theta) instead of cos (theta), and thus the solution is 573.6 kJ.

But my question is why is sin(theta) used instead of cos(theta) for the formula . I thought that if the direction the wagon is being pulled needs to be parallel to the horizontal component of the Force, which should be cos (theta)?

I'm just curious if anyone could help me see why W = F*d*sin(theta) is the appropriate formula for this example and not W = F*d*cos(theta).

I appreciate any response or feedback.
It's a mistake, You are correct. Cos(35º) should be used.

Also, note the data are given to only 1 or 2 significant figures. I'd round the answer to 2 significant figures.

The fact that the 'official' solution is both wrong and given to 4 significant figures isn't inspiring!

## What is work done on an object?

Work done on an object is the amount of energy transferred to the object when a force is applied to it and the object moves in the direction of the force.

## What does it mean to pull a wagon while lifting up at an angle?

Pulling a wagon while lifting up at an angle means that the force applied to the wagon is not in the same direction as the motion of the wagon. It involves both a horizontal and vertical component of force.

## How is the work done on an object calculated in this scenario?

The work done on an object in this scenario is calculated by multiplying the force applied to the object by the distance the object moves in the direction of the force. This takes into account both the horizontal and vertical components of the force.

## Does lifting the wagon while pulling it require more work to be done?

Yes, lifting the wagon while pulling it requires more work to be done because it involves applying force in two different directions, which results in a greater amount of energy being transferred to the wagon.

## What is the relationship between the angle of lifting and the work done on the wagon?

The relationship between the angle of lifting and the work done on the wagon is that the work done will be greater when the angle of lifting is smaller. This is because a smaller angle results in a greater vertical component of force, which requires more energy to be transferred to the wagon.

• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
507
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K