Calculating Distance: 12 kg Object with 24 J Kinetic Energy and 6 N Force

  • Thread starter Thread starter Nimmy
  • Start date Start date
  • Tags Tags
    Hard
AI Thread Summary
A 12 kg object with 24 joules of kinetic energy will move against a net resisting force of 6 Newtons until it stops. The work done by the resisting force must equal the object's kinetic energy, leading to the conclusion that it will travel 4 meters before coming to a stop. This calculation assumes the resisting force remains constant throughout the motion. The discussion highlights the importance of formulating questions that have practical applications rather than those that are overly complex or theoretical. Overall, the object moves 4 meters before its kinetic energy is fully dissipated.
Nimmy
Messages
41
Reaction score
0
A 12 kg object initally has 24 joules of kinetic energy. How far will the object move against a net resisting force of 6 Newtons? :rolleyes:

(A) 2 m
(B) 4 m
(C) 12 m
(D) 16 m
(E) 24 m
 
Physics news on Phys.org
Nimmy, please show us your ideas. We can not help you otherwise.

Read this first : https://www.physicsforums.com/showthread.php?t=28
 
Last edited:
think its easy...

the mass is moving with a velocity 2m/s right?
the resisting force would so a work in such a way that the mass stops after moving a certain distance...so the work done by the resisting force should be cancelling out the mass's kinetic energy...the resisting force is constant throughout right?then the block would be moving 4 metres as then the 6 Newton force would be doing a work of -24 joules cancelling out the mass's kinetic energy...i think its right...what kind of a question is this...are u preparing for some entrance examination...nimmy try to bring in question which has more application...anyone can create high class questions which no one can even answer...so grab such questions which are made not for the sake of making questions...but ...i m just speaking out that's al...leave it
 
Last edited:
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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