Solve Standard Mech Q: Acceleration & Distance in 2 Seconds

  • Thread starter Thread starter discombobulated
  • Start date Start date
  • Tags Tags
    Standard
AI Thread Summary
A 0.5kg bag is being pulled across a smooth horizontal floor by a string with a tension of 0.75N. The only horizontal force acting on the bag is the tension, leading to the calculation of acceleration using the formula a = F/m. The resultant force is equal to the tension since there is no friction, resulting in an acceleration of 1.5 m/s². Given that the bag starts from rest, it will move a distance of 6 meters in 2 seconds. The discussion clarifies the importance of recognizing the absence of friction in this scenario.
discombobulated
Messages
41
Reaction score
0
hi this is a simple question but for some reason it's not clicking.

A ball of mass 0.5kg is pulled across a smooth horizontal floor by a horizontal string attached to the dog. The tension in the string is 0.75N. Calculate the acceleration produced. Given the ball starts from rest, determine how far it will move in 2 seconds.

I know how to do the second part but some help would be appreciated on the first bit please.

F resultant = F - T
F-0.75 = ma
a = F- 0.75 /0.5
errm...and then?
 
Physics news on Phys.org
Is the ball rolling? or is it dragging across the floor? Some coeffeciant of friction would be needed if it was being dragged.
 
oops sorry i didn't read the question properly, it's actually a bag and it's being pulled across.
 
Since the floor is smooth, I would assume that the only horizontal force acting on the bag is the tension in the string, which is given.
 
Of course, because then the acceleration = 1.5 which is correct.

Thanks!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top