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The mass of lift is $2000 \mathrm{~kg}$. When the tension in the supporting cable is $28000 \mathrm{~N}$, then its acceleration is :
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1806 Upvotes
Verified Answer
The correct answer is:
$4 \mathrm{~ms}^{-2}$ upwards
$$
\begin{aligned}
& \Sigma F=m a \\
& \Rightarrow \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\
& \mathrm{a}=\frac{\mathrm{T}-\mathrm{mg}}{\mathrm{m}}=4 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
\begin{aligned}
& \Sigma F=m a \\
& \Rightarrow \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\
& \mathrm{a}=\frac{\mathrm{T}-\mathrm{mg}}{\mathrm{m}}=4 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
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