Please could you check this working for me?

  • Thread starter Thread starter lioric
  • Start date Start date
AI Thread Summary
The discussion focuses on a physics problem involving two blocks connected by a string over a frictionless pulley. Block A, with a mass of 50 kg, moves up an incline while block B, with a mass of 100 kg, drops down. The coefficient of kinetic friction is given as 0.25, and the user initially misapplies the forces acting on block A. After clarification, it is confirmed that the tension must account for the weight component acting opposite to it, leading to the correct formulation of the equations. The user expresses gratitude for the learning experience and acknowledges the correction.
lioric
Messages
335
Reaction score
26
Plz could you check this working for me??

Homework Statement


Two blocks, A and B (with mass 50 kg and 100 kg, respectively), are connected by a string, as
shown in figure.

The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between
block A and the incline is μk = 0.25. Determine the change in the kinetic energy of block A as
it moves from (C) to (D), a distance of 20 m up the incline (and block B drops down a
distance of 20 m) if the system starts from rest.



Homework Equations



Block A
T-μR+mg sinθ= ma

Block B

mg-T=ma




The Attempt at a Solution




acceleration in both blocks is the same since they are connected directly with no elastic strings

Block A
T=ma+μR-mg sinθ

Block B
T=mg-ma

Substitute for T

mg-ma=ma+μR-mg sinθ
(100 x 9.8) - (100a) = (50a) + (0.25 x 50 x 9.8 x cos 37) - (50 x 9.8 x sin 37)
980-100a-50a=97.83-294.89
-100a-50a=97.83-294.89-980
-150a=-1177.057
a=7.8


is this correct?
 

Attachments

  • Untitled.png
    Untitled.png
    27.9 KB · Views: 1,911
Physics news on Phys.org


lioric said:

Homework Statement


Two blocks, A and B (with mass 50 kg and 100 kg, respectively), are connected by a string, as
shown in figure.

The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between
block A and the incline is μk = 0.25. Determine the change in the kinetic energy of block A as
it moves from (C) to (D), a distance of 20 m up the incline (and block B drops down a
distance of 20 m) if the system starts from rest.



Homework Equations



Block A
T-μR+mg sinθ= ma
You have a signage error here...which way does the weight component act along the plane? Once you find the acceleration, then you need to find the speed of the blocks to get the KE change. Alternatively, using work-energy methods, you will get the same result.
 


oh i see so are you saying that it should be more like

T-μR-mg sinθ= ma

because the component of weight is opposite to tension?
 


lioric said:
oh i see so are you saying that it should be more like

T-μR-mg sinθ= ma

because the component of weight is opposite to tension?
Yes, that is correct, and i think the best way to proceed to then find the speed of the block A and its KE change.
 


Thank you very much i feel that i have learned a lot since i joined here
 
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