Help With Physics problems i'm having trouble

AI Thread Summary
The discussion centers on two physics problems involving motion and forces. The first problem requires calculating the stopping distance of a car on a slope, but the user struggles due to missing data, particularly the coefficient of kinetic friction (mu_k). For the second problem, which involves a mass revolving in a circle, the user seeks to determine the angle of the string from the vertical but is unsure about the correct approach to find it. The conversation highlights the importance of identifying all necessary variables and using appropriate formulas, such as centripetal acceleration for the second problem. Overall, the user is looking for guidance on solving these physics challenges effectively.
songokou77
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1)Find the stopping distance of that same car when it is traveling up a 17.1° slope, and it locks its wheels while traveling at 34.9 m/s (78.2 mi/hr). Assume that mu_k does not depend on the speed.


2)A mass of 4.100 kg is suspended from a 1.430 m long string. It revolves in a horizontal circle.
The tangential speed of the mass is 2.899 m/s. Calculate the angle between the string and the vertical (in degrees).



I'm having trouble with this two problems. The first one all i think of are the formulas : a=v^2/(2(distance), and i have to find the distance, but i don't understand :Assume ...
In the second problem i thought i could get the angle by find the inverse cosine of the ropes lenght, but I am not that sure
 
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songokou77 said:
1)Find the stopping distance of that same car when it is traveling up a 17.1° slope, and it locks its wheels while traveling at 34.9 m/s (78.2 mi/hr). Assume that mu_k does not depend on the speed.


2)A mass of 4.100 kg is suspended from a 1.430 m long string. It revolves in a horizontal circle.
The tangential speed of the mass is 2.899 m/s. Calculate the angle between the string and the vertical (in degrees).

Step 1: What is the data I have?
v(i) = 34.9m/s
v(f)
mg(y) = 9.8m/s^2
mg(parrallell) ' to slope = 9.8*sin(17.1)
mg(perpendicular) = 9.8*cos(17.1)

Alright, what don't we know?
mu_k
t
d

Uh oh, without those 3 we can't solve mu_k. That means we can't find the friction. You're not telling me all the data.

2. You'll need the formula for centripedal acceleration. a = v^2/r
 
Actually this was a two part problem the first part was:
A 680.0 kg car traveling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft). Find the size of the horizontal force which the car applies on the road while stopping.
The answer for that one was : 4.06E+03 N
 
songokou77 said:
Actually this was a two part problem the first part was:
A 680.0 kg car traveling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft). Find the size of the horizontal force which the car applies on the road while stopping.
The answer for that one was : 4.06E+03 N

Find mu_k with the first question and the second becomes possible.
 
I used the data I have, but either the formula I'm using is incorrect or I'm doing something wrong. I tried using a=V^2/2(d) and i tried to solve for d there but i don't get a correct answer. Can you recommend any formula for this maybe a variation of the 3rd law( i assume this is a 3rd law problem).
 
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