Projectile problem: with coefficient of kinetic friction?

AI Thread Summary
Julia needs to determine the optimal horizontal distance to drop a newspaper while driving at 15 m/s from a height of 1.00 m. The calculation reveals that the time it takes for the newspaper to fall is approximately 0.45 seconds, resulting in a horizontal distance of 6.78 m before it hits the ground. After landing, the newspaper slides due to kinetic friction, which is calculated using a coefficient of 0.40, leading to a stopping distance of about 28.7 m. The discussion highlights that while the calculated distance is correct, real-world factors like air resistance and surface roughness may affect the actual sliding distance. Overall, the problem illustrates the two phases of motion: projectile and sliding.
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top