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

[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.

 

[Steve4Physics]

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!