Calculating Mass and Normal Force of the Most Massive Car Ever Built

AI Thread Summary
The discussion revolves around calculating the mass and normal force of the most massive car ever built, which was the official car of the General Secretary of the Communist Party in the former Soviet Union. The car is moving down a 10-degree slope, and the net force acting on it when the brakes are applied is -2.00 x 10^4 N, with a coefficient of kinetic friction of 0.797. Participants discuss the appropriate equations to use, including Newton's second law, and the importance of a free-body diagram. One user suggests that the mass can be calculated using the formula F/g = mass, which provides clarity to others in the thread. The conversation highlights the challenges of applying physics concepts to real-world scenarios.
Twilit_Truth
Messages
34
Reaction score
0
This is my first time here, so if I put this in the wrong place I am sorry. I will place the entire problem here, exactly as it appears.

The most massive car ever built was the official car of the General Secritary of the Communist Party in the former Soviet Union. Suppose this car is moving down a 10.0 degree slope when the driver suddenly applies the brakes. The net force acting on the car as it stops is -2.00 X 10^4 N. If the coefficient of kinetic friction between the car's tires and the pavement is 0.797, what is the car's mass? What is themagnitude of the normal force that the pavement exerts on the car?


I have no clue as to what equations to use. Sorry.



I made a free-body diagram, but I think I messed it up.

Thank you for your time.
 
Last edited:
Physics news on Phys.org
Can you explain all of the forces that are present on your diagram? What's Newton's second law?
 
  • Like
Likes Simone Eissa
Twilit_Truth said:
This is my first time here, so if I put this in the wrong place I am sorry. I will place the entire problem here, exactly as it appears.

The most massive car ever built was the official car of the General Secritary of the Communist Party in the former Soviet Union. Suppose this car is moving down a 10.0 degree slope when the driver suddenly applies the brakes. The net force acting on the car as it stops is -2.00 X 10^4 N. If the coefficient of kinetic friction between the car's tires and the pavement is 0.797, what is the car's mass? What is themagnitude of the normal force that the pavement exerts on the car?


I have no clue as to what equations to use. Sorry.



I made a free-body diagram, but I think I messed it up.

Thank you for your time.

Just asking someone who knows, couldn't you find the mass by doing F/g = mass?
 
You could. Thank you SO much for that comment. Apparently, I'm brain-dead, lol.
 
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