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# A 15 kg box sits on a table. The coefﬁcient of static friction $\mu_s$ between table and box is 0.3, and the coefﬁcient of kinetic friction $\mu_k$ is 0.2.(a) What is the force required to start the box moving?

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### Given

We are given the mass of the box by $m$ = 15 kg and the coefficient of static friction is $\mu_s$ = 0.3 and the coefficient of kinetic friction $\mu_k$ = 0.2

### Solution

The static means, the box doesn't move and the friction force between the table and the box holds the box. Hence, to just start moving the box we should apply a force equals to this friction force, so it is given by

$\begin{equation} F = f = \mu_s mg \end{equation}$

Where $m$ is the mass of the box and $g$ is the gravitational acceleration and equals $9.8 \mathrm{~m/s^2}$. Now we can plug our values for $\mu_s$, $m$ and $g$ into equation (1) to get the force required to start the moving

\begin{align*} F = \mu_s mg = (0.3)(15\,\text{kg})(9.8 \mathrm{~m/s^2}) = \boxed{44.1 \,\text{N}} \end{align*}

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