Projectile problem: with coefficient of kinetic friction?

Click For Summary

Homework Help Overview

The problem involves a scenario where a newspaper is dropped from a moving vehicle, requiring the calculation of the horizontal distance it travels before coming to rest, factoring in the coefficient of kinetic friction between the newspaper and the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the two stages of motion: the projectile phase and the sliding phase on the ground. There is uncertainty about how to incorporate the coefficient of kinetic friction into the calculations. Some participants express confusion regarding the implications of the coefficient of friction on the distance traveled before stopping.

Discussion Status

Some participants have provided guidance on breaking down the problem into two stages and have confirmed calculations related to the sliding distance. There is acknowledgment of the potential discrepancies between theoretical calculations and real-world scenarios, particularly regarding the coefficient of friction.

Contextual Notes

Participants note that the coefficient of kinetic friction may be low for a rough surface like a road, and air resistance is ignored in the calculations, which could affect the realism of the results.

paxian
Messages
14
Reaction score
0

Homework Statement


Julia is delivering newspapers. Suppose she is driving at 15 m/s along a straight road and wants to drop a paper out the window from a height of 1.00m so it slides along the shoulder and comes to rest in the customer's driveway. At what horizontal distance before the driveway should she drop the paper? The coefficient of kinetic friction between the newspaper and the ground is 0.40. Ignore air resistance and assume no bouncing and rolling.

Homework Equations



delta y = Viy* delta t + 1/2 ay* delta t^2
delta x= Vix * detal t

The Attempt at a Solution


delta y = Viy* delta t + 1/2 ay* delta t^2
-1m=0 + 1/2(-9.8m/s^2)*delta t^2
==> delta t = sq. root of (-2* -1m/-9.8m/s^2) = 0.45s
delta x= Vix * detal t=15m/s * 0.45s = 6.78m

I am not sure what to do with the coefficient of kinetic friction between the newspaper and the ground? Please help! Or is it a trick to throw me off... the answer is just simply 6.78 m?
 
Physics news on Phys.org
There are two stages of the motion. First the newspaper is a projectile, with horizontal initial velocity. In the second stage, it slides on the ground. Find the distance traveled in the horizontal direction as a projectile and then the distance traveled to full stop.

ehild
 
ehild said:
There are two stages of the motion. First the newspaper is a projectile, with horizontal initial velocity. In the second stage, it slides on the ground. Find the distance traveled in the horizontal direction as a projectile and then the distance traveled to full stop.

ehild

Thank you so much.. NOw it makes sense why they thro that coefficient of K in there. soFriction the only force working against the newspaper at that phase?!

net force=ma = - Friction force = ma
- (coefficient k * m* g) = ma
a = -3.92
vf^2 = vi^2 + 2a delta x
x= 28.7m?

the distance from second stage is kinda high, it took 28.7m to reach full stop?
that's kinda crazy! Did I do it right? Thanks in advance!
 
That 28.7 m is correct.
I think that the newspaper will not slide that long in reality, but that coefficient of friction is rather low for a rough surface as the road, and air resistance also acts on the newspaper. ehild
 
ehild said:
That 28.7 m is correct.
I think that the newspaper will not slide that long in reality, but that coefficient of friction is rather low for a rough surface as the road, and air resistance also acts on the newspaper.


ehild

Thanks so much! You're so helpful! Have a wonderful day :)
 
paxian said:
Thanks so much! You're so helpful! Have a wonderful day :)

Thank you:smile:

ehild
 

Similar threads

Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
18K
Replies
6
Views
2K
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
18
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
4K