Newton's Laws of Motion - Force Problem

Click For Summary
SUMMARY

The discussion centers on calculating the average force exerted by a target on a 0.075-kg arrow that penetrates 3.8 cm after hitting at a velocity of 21 m/s. The average force calculated is 435 N, directed opposite to the arrow's motion. Additionally, the discussion poses a scenario where an identical arrow strikes the target at 42 m/s, prompting the need to determine the new penetration depth under the same average force conditions.

PREREQUISITES
  • Newton's Second Law of Motion
  • Basic kinematics equations
  • Understanding of force and acceleration
  • Knowledge of unit conversions (e.g., meters to centimeters)
NEXT STEPS
  • Study the implications of Newton's Third Law of Motion in collision scenarios
  • Learn about energy conservation principles in elastic and inelastic collisions
  • Explore advanced kinematics involving variable forces
  • Investigate the relationship between impact velocity and penetration depth in materials
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of force calculations in motion scenarios.

Yshai24
Messages
6
Reaction score
0

Homework Statement


A 0.075-kg arrow hits the target at 21 m/s and penetrates 3.8cm before stopping. (a) What average force did the target exert on the arrow? (b) What average force did the arrow exert on the target? (c) An identical arrow strikes the target at 42m/s. If the target exerts the same average force as before, what's the penetration depth?

Known: mass: .075kg velocity (?) 21m/s, displacement: 3.8cm

Homework Equations


Here I am a little confused. (Or a lot confused)
Fnet=m x a

a=velocity final squared - velocity initial squared / 2(Δ)? Not sure if this equation applies

The Attempt at a Solution


I used the equation to figure out the acceleration since I have the mass. So here is where I am at:
a=0-(21m/s)2 / 2(.038m) = -5.8 x 10^3 m/s2

Then I used that to find Fnet=m x a = (.075kg)(-5.8 x 10^3m/s2)= -435N

I feel like I am on a completely wrong track and am not sure how to answer the question. Is the target applying a force of -435N?

Any help in the right direction is greatly appreciated, thank you for your time.

V/r
Jesse
 
Physics news on Phys.org
Looks good! Yes , if the arrow is moving left to right, the target applies a force on the arrow of 435 N to the left. What force and in what direction does the arrow apply on the target?
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
Replies
13
Views
3K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K