Figure out how far up the incline of the hill the box slides.

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SUMMARY

The problem involves calculating how far a box slides up an incline after being released with an initial speed of 15 m/s at a 30-degree angle. The correct acceleration due to gravity acting down the incline is determined using the formula a = (9.81 m/s²) * sin(30), resulting in a distance of 23 meters. The initial miscalculation used the cosine component instead of the sine, leading to an incorrect distance of 13.25 meters. The final solution confirms the importance of correctly identifying the components of gravitational force on an incline.

PREREQUISITES
  • Understanding of kinematic equations
  • Knowledge of trigonometric functions, specifically sine and cosine
  • Familiarity with gravitational acceleration (9.81 m/s²)
  • Basic concepts of motion on an incline
NEXT STEPS
  • Study kinematic equations in detail, focusing on motion under constant acceleration
  • Learn about forces acting on inclined planes and their components
  • Explore the application of trigonometric functions in physics problems
  • Practice similar problems involving motion on inclines to reinforce understanding
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and motion on inclined planes, as well as educators looking for examples of kinematic applications.

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Homework Statement


At the bottom of a hill, a box has a speed of 15 m/s. The hill forms an angle of 30 degrees with the horizontal. How far up the incline of the hill does the box slide?

Homework Equations


My answer:
d = (vf2 + vi2) / 2a

The Attempt at a Solution


vi = 15 m/s
vf = 0 m/s

a = (-9.81 m/s2)*cos(30) = -8.49 m/s2

d = 13.25 m

However, the answer key says d = 23 m. This is the answer if you use a = (9.81 m/s^2)*sin(30) instead. Help?
 
Physics news on Phys.org
What's the component of gravity acting down the incline? That's where sinθ comes in.
 
Oops, I had 30 degrees in the wrong corner of the triangle. It works now! Thanks.
 

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