Question
The two blocks in an Atwood machine have masses $2.0kg$ and $3.0kg$. Find the work done by gravity during the fourth second after the system is released from rest.
✓
Answer
Given, $m_1 = 3 kg, m_2 = 2kg$, t = during $4^{th}$ second From the freebody diagram,
T - 3g + 3a = 0 ...(i) T - 2g - 2a = 0 ...(ii) Equation (i) & (ii), we get 3g – 3a = 2g + 2a $\Rightarrow\text{a}=\frac{\text{g}}{5}\text{m}/\text{sec}^2$ Distance travelled in $4^{th}$ sec is given by, $\text{S}_{4\text{th}}=\frac{\text{a}}{2}(2\text{n}-1)$ $=\frac{\Big(\frac{\text{g}}{5}\Big)}{\text{S}}(2\times4-1)$
$=\frac{7\text{g}}{10}=\frac{7\times9.8}{10}\text{m}$ Net mass $‘m’ = m_1 - m_2 = 3 - 2 = 1kg$ So, decrease in P.E. = mgh $=1\times9.8\times\frac{7}{10}\times9.8$
$=67.2=67\text{J}$
T - 3g + 3a = 0 ...(i) T - 2g - 2a = 0 ...(ii) Equation (i) & (ii), we get 3g – 3a = 2g + 2a $\Rightarrow\text{a}=\frac{\text{g}}{5}\text{m}/\text{sec}^2$ Distance travelled in $4^{th}$ sec is given by, $\text{S}_{4\text{th}}=\frac{\text{a}}{2}(2\text{n}-1)$ $=\frac{\Big(\frac{\text{g}}{5}\Big)}{\text{S}}(2\times4-1)$
$=\frac{7\text{g}}{10}=\frac{7\times9.8}{10}\text{m}$ Net mass $‘m’ = m_1 - m_2 = 3 - 2 = 1kg$ So, decrease in P.E. = mgh $=1\times9.8\times\frac{7}{10}\times9.8$
$=67.2=67\text{J}$
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