Average Resistance Force Question

Click For Summary
SUMMARY

The average resistance force exerted by sand on a 5.2 g ball bearing, fired vertically downward from a height of 18 m with an initial speed of 14 m/s and burying itself 21 cm into the sand, is calculated to be -6.81 N. To determine this force, one can utilize either the constant acceleration equations or the conservation of energy principles. The key equations involve calculating the final velocity upon impact and applying Newton's second law (F = ma) or the energy conservation formula.

PREREQUISITES
  • Understanding of kinematics and constant acceleration equations
  • Familiarity with Newton's second law (F = ma)
  • Knowledge of conservation of energy principles
  • Basic proficiency in algebra and physics problem-solving
NEXT STEPS
  • Study the kinematic equations for constant acceleration
  • Learn about the conservation of mechanical energy in physics
  • Explore examples of calculating forces using F = ma
  • Investigate real-world applications of resistance forces in physics
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone interested in understanding forces and energy conservation in motion scenarios.

alexandray77
Messages
2
Reaction score
0
Our question to solve is:
A 5.2 g Ball bearing is fired vertically downward from 18m with an initial speed of 14 m/s. It buries itself a depth of 21 cm in the sand. What average resistance force does the sand exert on the ball bearing?

We know the answer is -6.81N but we are having a hard time figuring out which equations to use to solve the problem. Please help! Thanks!
 
Physics news on Phys.org
You can use either constant acceleration formulae or conservation of energy to solve this problem.

With constant acceleration you are going to want to calculate the velocity of the ball at the instant it hits the ground. You will then have initial and final velocity and displacement. You can solve for a, and use F = ma to solve for F.With energy, you can use: [tex]\frac{1}{2}mv_1^2 + mg\Delta h_1 + F_1\Delta d_1 = \frac{1}{2}mv_2^2 + mg\Delta h_1 + F_2\Delta d_2[/tex]
some of these terms will be zero, you are solving for F2
 
thank you!
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K