2 problems: potential erg & tension/friction

  • Thread starter Thread starter Falinox
  • Start date Start date
  • Tags Tags
    Potential
AI Thread Summary
The discussion revolves around two physics problems related to potential energy and tension/friction. Participants verify calculations and provide corrections, noting that the initial velocity for problem 7 should be 4.76 m/s. One user acknowledges a mistake in their calculations for problem 7, realizing they mismanaged a factor of 2. For problem 5, the acceleration was confirmed to be approximately 0.57 m/s², aligning with another user's calculation of 0.56645 m/s². Overall, the thread emphasizes collaborative problem-solving and verification of physics concepts.
Falinox
Messages
11
Reaction score
0

Homework Statement


Question1.jpg

http://i754.photobucket.com/albums/xx183/reddiesel08/Question1.jpg
Question2.jpg

http://i754.photobucket.com/albums/xx183/reddiesel08/Question2.jpg

2. The attempt at a solution
Attempt2.jpg

http://i754.photobucket.com/albums/xx183/reddiesel08/Attempt2.jpg
Attempt.jpg

http://i754.photobucket.com/albums/xx183/reddiesel08/Attempt.jpg

Need someone to check my work and help solve these two problems, thanks in advance to anyone willing to help.
 
Last edited by a moderator:
Physics news on Phys.org
5 is correct, but I did not check the numbers for this one.

7, your method is correct, but your final answer is off. My calculator gives me vi = 4.76 m/s
 
rock.freak667 said:
5 is correct, but I did not check the numbers for this one.

7, your method is correct, but your final answer is off. My calculator gives me vi = 4.76 m/s

Thank you! My mistake on question 7 was I forgot to multiply the 9.8(0.68) by 2 when I moved the 1/2 from the left side to the right side of the equation. I also came up with the answer 4.76 m/s.

On #5 I got the answer a = .57 m/s squared can someone verify this for me.
 
a = 0.56645 m/s^2, so you are correct.
 
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