## Related questions with answers

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?

Solution

Verified### 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

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

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick#### Matter and Interactions

4th Edition•ISBN: 9781118875865 (2 more)Bruce A. Sherwood, Ruth W. Chabay## More related questions

1/4

1/7