A question on work, power and energy

  • Thread starter Thread starter ritik.dutta3
  • Start date Start date
  • Tags Tags
    Energy Power Work
AI Thread Summary
The discussion revolves around calculating the work done by gravity in an Atwood machine with masses of 2 kg and 3 kg. The initial calculations incorrectly used the total distance traveled over four seconds instead of focusing on the distance during the specific fourth second. The correct approach involves determining the distance traveled between the third and fourth seconds to find the work done. The user realizes the mistake after clarification, leading to an understanding of the need for precise time intervals in such calculations. This highlights the importance of careful analysis in physics problems.
ritik.dutta3
Messages
8
Reaction score
0

Homework Statement


The two blocks in an Atwood machine have masses 2 kg and 3 kg. Find the work done by gravity during the fourth second after the system is released from rest.


Homework Equations


Work done by gravity= mgh



The Attempt at a Solution


Both the masses will have the same acceleration 'a'. Let T be the tension in the string.
m1= 2 kg, m2= 3 kg, g= 10 m/s^2

T-m1g=m1a
therefore, T- 20= 2a

m2g- T= m2a
therefore, 30 - T= 3a
on solving both the equations, a= 2 m/s^2

distance traveled by the blocks would be 1/2(at^2). So, the heavier block would travel 16 m downwards and the lighter block 16 m upwards.
Net mass= 3-2= 1 kg
hence, the work done should be
W= mgh= 1 x 10 x 16 = 160J

but the answer is 67J.
where am I going wrong?
 
Physics news on Phys.org
ritik.dutta3 said:
distance traveled by the blocks would be 1/2(at^2). So, the heavier block would travel 16 m downwards and the lighter block 16 m upwards.

You have used distance traveled in 4 seconds .Whereas you need distance traveled during the 4th second i.e between 3rd and 4th sec.
 
  • Like
Likes 1 person
Got it! Thanks! :D
 
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