Question

The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is 16.0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

HEY!!

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⚫2mg−T=2ma  -----------(1)

⚫T−mg=ma  -----------(2)

⚫from (1) and (2),

⚫a= g/3

⚫y=ut+1/2at^2

⚫y=0× 1+ 1/2× g/3 (1)^2

⚫y= 10/6= 5/3 m  .....[as g=10 m/s^2]

⚫Total loss of P.E, U=2mgy−mgy

⚫U=mgy

⚫U=m×10×5/3

⚫U= 50/3m

✔✔Where m=mass, which must be know

Answer 2

Given :-
Mass M = 2 m.
Tension T = 16.0 N
Now assume acceleration of the block is 'a '
Then from the equation of the motion for the block
Mg - T = Ma
T = Mg - Ma
T = M ( g- a)
T = 2m (g-a) ------→ eq. (i)

and also, For second block
T - mg = ma 
T = m( a +g)  ------→eq (ii)

Now equating both equation we get,
∴ 2m(g-a) = m( g+a)
2g -2a = g + a
g = 3a
or a = g/3

Now we know the formula s= ut + 1/2 (at²)
s = 0 + 1/2 [ (g/3) × 1² ]
s = g/6 .

For calculating the potential energy need to know mass. 
from eq . (ii)
T = m (g +a )
T = m ( g + g/3)
or m = T / [g + (g/3)]
m = 3T / 4g

we know that Mass M = 2m
M = 6T /4g

Thus total decrees in Gravitational potential energy = Mgs - mgs
=Mgs - mgs
=(6T/4g) gs - (3T/4g) gs
= [(6T/4g) g × g/6 ] - [(3T/4g) g × g/6]    {as, s = g/6}
=  Tg/4 - Tg/8
= Tg/8
= (16 × 9.8) / 8
= 156.8 /8
P.E= 19.6 J.

Hence decrease in P.E is 19.6 J


Hope it Helps.