Please could you check this working for me?

  • Thread starter lioric
  • Start date
In summary, the pulley has a coefficient of kinetic friction of 0.25, and as block A moves up the incline, its kinetic energy decreases by 97.83 joules.
  • #1
lioric
306
20
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?
 

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  • #2


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.
 
  • #3


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?
 
  • #4


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.
 
  • #5


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

FAQ: Please could you check this working for me?

1. What does "checking the working" mean in the context of your work?

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