Minimum stopping distance problem

AI Thread Summary
The discussion revolves around calculating the minimum stopping distance for a 6000 kg steel load on a truck traveling at 20 m/s, held by brackets exerting a maximum force of 8000 N. The acceleration calculated is -1.33 m/s², leading to a stopping time of 15 seconds. Using the kinematic equation, the minimum stopping distance is determined to be 150 meters. The user expresses gratitude for the assistance received in confirming their calculations. The problem highlights the importance of understanding forces and motion in physics.
Gashouse
Messages
10
Reaction score
0
A load of steel of mass 6000kg rests on the flatbed of a truck. It is held in place by metal brackets that can exert a maximum horizontal force of 8000 N. When the truck is traveling 20 m/s, what is the minimum stopping distance if the load is not to slide forward into the cab?
 
Physics news on Phys.org
ok here how i did it,...im gettin an answer of 150m

first find the acceleration in the negative x direction the brackets can exert which is.
8000=(6000)a

a= -1.33


then find the time it takes to stop using v=vo +at

0= 20 + (-1.33) t t= 15sec


then find distance using x=xo +vot+.5at^2



x = 20(15) + .5(-1.33)(15)^2

x=150m
 
Thanks PN

Hey PHYSICS NOOB, I really appreciate the help. I worked the problem before coming on the forum and did'nt think that I was doing it correct. Thanks in advance again and for the fast response.
 
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