Question

The heavier block in an Atwood machine has 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.

Answer 1

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19.6 joule

Answer 2

The decrease in P.E is  19.6 J

Explanation:

Step 1:

Given values in the question  

Mass M = 2 m.

Tension T = 16.0 N

Step 2:

Now assume that the block's acceleration is ' a ' then from the block's motion equation

Mg - T = Ma

T = Mg - Ma

T = M (g- a)

T = 2m (g-a) ------ eqn (i)

Step 3:

And for the second block as well

T - mg = ma  

T = m (a +g)  ------eqn (ii)

Now we're using the same equation,

2m(g-a) = m (g+a)

2g - 2a = g + a

g = 3a

$\text { or } a= \frac{g}{3}$

Step 4:

Now we know the formulas  $=u t+\frac{1}{2} a t^{2}$

$\begin{aligned}&s=u t+\frac{1}{2} \times \frac{g}{3} 1^{2}\\&s=0+\frac{1}{2} \times \frac{g}{3}\end{aligned}$

$s=\frac{g}{6}$

Step 5:

Mass needs to be known to measure the potential energy.

from eq . (ii)

$\begin{aligned}&\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\\&\mathrm{T}=\mathrm{m}\left(\mathrm{g}+\frac{g}{3}\right)\\&m=\frac{T}{g+\frac{g}{g}}\\&m=\frac{T}{\frac{3 g+g}{3}}\\&m=\frac{3 T}{4 g}\end{aligned}$

we know that  

Mass M = 2m

$\begin{aligned}&M=2 \times \frac{3 T}{4 g}\\&m=\frac{6 T}{4 g}\end{aligned}$

Step 6:

Full decrease in the energy potential of gravity = Mgs - mgs

= Mgs – mgs

$\begin{aligned}&=\frac{6 T}{4 g} \times g s-\frac{3 T}{4 g} \times g s\\&=\frac{6 T}{4 g} \times g \times \frac{g}{6}-\frac{3 T}{4 g} \times g \times \frac{g}{6}\end{aligned}$

[as, $s =\frac{g}{6}$ ]

$\begin{aligned}&=\frac{T}{4} \times g-\frac{T}{8} \times g\\&=\frac{T}{8} g\\&G=9.8\\&=\frac{16}{8} \times 9.8\\&=\frac{156.8}{8}\end{aligned}$

P.E= 19.6 J.

The decrease in P.E is therefore 19.6 J