(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

Force = mass * acceleration

Fg = ma

Vf^2= 2ad + Vi^2

3. 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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# Homework Help: How fast is the wagon going after moving 82.1 m up the hill?

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