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A block of mass $5 \mathrm{~kg}$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is 0.1 , then the force needed to accelerate the block to $3 \mathrm{~m} / \mathrm{s}^2$ to the right is close to
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Verified Answer
The correct answer is:
$22 \mathrm{~N}$
Component of force accelerating the block is $f \cos \theta$.
Net force on block, $F_{\text {net }}=F \cos \theta-\mu N$
$$
\begin{aligned}
& =F \cos 30^{\circ}-0.1 \times 5 \times 9.8 \\
& =\frac{\sqrt{3} F}{2}-49
\end{aligned}
$$
From, $F_{\text {net }}=m a$, we have,
$$
\frac{\sqrt{3}}{2} F-4.9=5 \times 3 \Rightarrow F=\frac{19.9}{\sqrt{3}}=22.9 \mathrm{~N}
$$
This is close set to option (b).
Net force on block, $F_{\text {net }}=F \cos \theta-\mu N$
$$
\begin{aligned}
& =F \cos 30^{\circ}-0.1 \times 5 \times 9.8 \\
& =\frac{\sqrt{3} F}{2}-49
\end{aligned}
$$
From, $F_{\text {net }}=m a$, we have,
$$
\frac{\sqrt{3}}{2} F-4.9=5 \times 3 \Rightarrow F=\frac{19.9}{\sqrt{3}}=22.9 \mathrm{~N}
$$
This is close set to option (b).
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