Work done by Tension Ques: Friction 0.5n

AI Thread Summary
The discussion focuses on calculating the speed of a block sliding down an incline after moving 100 cm, using energy conservation principles. A frictional force of 0.5 N is acting on the block, raising the question of whether the work done by tension should be included in the calculations. The total energy before and after the block moves is set to zero, leading to the equation that balances force times distance, kinetic energy, and potential energy. The vertical distance fallen, denoted as h, is also a critical factor in the energy equation. Ultimately, the role of tension in doing work is a key point of inquiry in the energy conservation analysis.
naman0804
Messages
2
Reaction score
0
ques: how fast block moves after sliding 100 cm on incline?[by energy conservation]
A frictional force of 0.5 n actsdoubt: should work done by tension be accounted?
 

Attachments

  • DSC05223.jpg
    DSC05223.jpg
    22.5 KB · Views: 520
Physics news on Phys.org
Did you give us all the information that you were given?
 
yes i did ... i just want to know if tension does work
 
Before the block moves you can set the total energy to zero, after the block has moved 1m the total energy is still zero and is,

0 = .5N X 1m + mV^2/2 - mgh = force times distance + kinetic energy - potential energy

where h is the vertical distance the block falls.
 
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