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

  • Thread starter Thread starter missashley
  • Start date Start date
  • Tags Tags
    Hill
Click For Summary
SUMMARY

The discussion centers on calculating the final velocity of a 40.5 kg wagon towed up a 16.5-degree incline after moving 82.1 meters. The tow rope exerts a force of 125 N, and the gravitational force is calculated as 397.305 N. The correct approach involves resolving the gravitational force into components, with the force acting down the slope being mg.sin(16.5 degrees). The final velocity calculated was 32.213 m/s, but the initial calculations contained a trigonometric error regarding the direction of gravitational force components.

PREREQUISITES
  • Understanding of Newton's Second Law (Force = mass * acceleration)
  • Knowledge of trigonometric functions (sine and cosine)
  • Familiarity with kinematic equations (Vf^2 = 2ad + Vi^2)
  • Basic concepts of forces on inclined planes
NEXT STEPS
  • Review the application of trigonometric functions in physics problems
  • Study the effects of forces on inclined planes in detail
  • Learn about kinematic equations and their applications in motion analysis
  • Explore examples of force resolution in two dimensions
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined plane problems, as well as educators seeking to clarify concepts related to forces and motion.

missashley
Messages
34
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

Replies
3
Views
5K
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K