How fast is the wagon going after moving 82.1 m up the hill?

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A 40.5 kg wagon is towed up a 16.5-degree incline with a force of 125 N, starting from rest and moving 82.1 m. The calculations initially miscalculated the gravitational force components, leading to an incorrect acceleration. The correct gravitational force down the slope should be calculated using mg.sin(16.5 degrees), which is essential for determining the net force acting on the wagon. After correcting the trigonometric error, the net force indicates that the wagon cannot move up the slope as initially assumed. The discussion emphasizes the importance of accurate trigonometric calculations in physics problems involving inclined planes.
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Homework Statement



A 40.5 kg wagon is towed up a hill inclined at 16.5 degrees with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 125 N on the wagon. Assume that the wagon starts from rest at the bottom of the hill, and disregard friction and significant figures.

The acceleration of gravity is 9.81 m/s^2

How fast is the wagon going after moving 82.1 m up the hill? Answer in m/s

Homework Equations



Force = mass * acceleration
Fg = ma
Vf^2= 2ad + Vi^2


The Attempt at a Solution



40.5 kg * 9.81 m/s^s = 397.305 kg/m/s^2 or 397.305 N

Tilting the time of reference, so that you just need to find the Fg of the diagram.
Fg in the x direction = 397.305 cos 16.5 = 380.8438748
Fg in the y direction = 397.305 sin 16.5 = 112.8407165

Just adding the forces in the x direction
380.8438749 - 125 = 255.9438748
255.9438748 / 40.5 kg = 6.319601847 = acceleration in the x direction

Acceleration in the y direction = 0 because the wagon isn't moving up and down

Vf^2 = 2ad + Vi^2
Vf^2 = 2 (6.3196017847)(82.1) + 0 ^2
Vf^2 = square root of 1037.678623
Vf = 32.21301947 m/s

I got 32.21301947 m/s as my answer

Did I do anything wrong?
 
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Fg in the x direction = 397.305 cos 16.5 = 380.8438748
Fg in the y direction = 397.305 sin 16.5 = 112.8407165
You have it backwards here. Fg in the x direction should be 397.305 sin 16.5 = 112.84. Check your trig. Other than that, I think you've got the right idea.
 
You have the force acting down the slope as 380.8 N, with 120 N acting up the slope, from the tow rope.

That leaves you with 260.8 N down the slope moving the wagon - which means the wagon is going to move backwards.

But the wagon is already at the bottom of the slope! It can't move backwards anymore.

This should have pointed out the error in your work.

The actual error is a trig one.

The component of the wagon's weight down the slope is mg.sin(alpha) - where here alpha = 16.5 degrees

Apart from that, your working is fine.
 
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