Question

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.

Answer 1

Work done by gravity in 4th sec = 70 J

Explanation:

m1 = 3 kg

m2 = 2kg

t = 4 seconds

From the freebody diagram, we infer that:

T - 3g + 3a = 0 -------(1)

T - 2g - 2a = 0 -------(2)

From the above equations, we get that:

3g - 3a = 2g + 2a

Therefore a = g/5 ms-2

Distance travelled in 4th second is:

S (4th) = a/2 (2n-1) = (g/5) / s * (2*4-1) = 7g/10 = 7 *10 / 10 (assuming g = 10 ms-2)

S (4th) = 7 meters

Net mass m = m1 - m2 = 3 - 2 = 1 kg

Work done by g = mgh = 1 * 10 * 7 = 70 J

Answer 2

The work done by gravity, is 67.28 J.

Explanation:

Step 1:

Given values in the questions  

Here are the Masses given

M =  3 kg .

m = 2 kg.

For the T-string stress.  

Stress in the string is the sum of the force due to the block's motion of gravity and energy.

T = ma + mg  

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

Similarly,

T = M (g - a)------------eqn (ii)

Step 2:

Now we're equating the two equations,

mg + ma = Mg - Ma

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

$\begin{aligned}&a=\frac{(M-m) g}{(M+m)}\\&a=\frac{(3-2) \mathrm{g}}{3+2}\\&a=\frac{E}{5} m / s^{2}\end{aligned}$

Step 3:

Using the motion equation

$S=u t+\frac{1}{2} a(2 t-1)$

Where t = 4 seconds, u = 0, [Because the block is at rest]

$\begin{aligned}&S=0+\frac{1}{2} \frac{g}{5}(2 \times 4-1)\\&S=\frac{g}{10}(8-1)\\&S=\frac{g}{10}(7)\\&S=\frac{7 \mathrm{g}}{10} \mathrm{m}\end{aligned}$

Step 4:

The work done by gravity is therefore:-

=Mgs - mgs

=gS (M-m)

$\begin{aligned}&=\frac{g\left(\frac{7 g}{10}\right)}{3-2}\\&=\frac{7 \mathrm{s}^{2}}{10}\\&G=9.8\\&=\frac{7 \times 9.8^{2}}{10}\\&=\frac{7 \times 96.04}{10}\\&=\frac{672,28}{10}\\&=67.28 \mathrm{J}\end{aligned}$

The work done by gravity, therefore, is 67.28 J.