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

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In summary, a 40.5 kg wagon is being towed up a 16.5 degree inclined hill by a force of 125 N. Assuming the wagon starts at rest and there is no friction, the wagon will have a velocity of 32.2 m/s after moving 82.1 m up the hill. However, there was an error in the calculation of the weight component, which should have been mg.sin(alpha).
  • #1
missashley
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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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  • #2
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.
 
  • #3
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.
 

1. How is the speed of the wagon calculated after moving 82.1 m up the hill?

The speed of an object is calculated by dividing the distance traveled by the time it took to travel that distance. In this case, the speed of the wagon can be calculated by dividing 82.1 m by the time it took to move up the hill.

2. What units are typically used to measure the speed of the wagon?

The speed of an object is commonly measured in meters per second (m/s) or kilometers per hour (km/h). These units represent the distance traveled per unit of time.

3. Can the speed of the wagon change during its motion up the hill?

Yes, the speed of the wagon can change during its motion up the hill. If the wagon encounters any obstacles or changes in the incline of the hill, its speed may increase or decrease.

4. How does the weight of the wagon affect its speed?

The weight of the wagon can impact its speed due to the force of gravity. A heavier wagon will require more force to move up the hill, potentially resulting in a slower speed compared to a lighter wagon.

5. Are there any factors besides distance and time that can affect the speed of the wagon?

Yes, there are other factors that can affect the speed of the wagon. These can include external forces such as friction, air resistance, and the incline of the hill. Additionally, the condition of the wagon and the amount of force applied to move it can also impact its speed.

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