# If the coefficient of kinetic friction between

If the coefficient of kinetic friction between the 50-kg crate and the ground is $\mu_k=0.3$, determine the distance the crate travels and its velocity when $t=3$ s. The crate starts from rest, and P = 200 N.

Image from: R. C. Hibbeler, K. B. Yap, and S. C. Fan, Mechanics for Engineers: Dynamics (SI Edition), 13th ed. Singapore: Pearson Education South Asia, 2013.

#### Solution:

Let us first draw a free body diagram. Remember that friction is always opposite to the direction the object moves.

$N_f$ is the normal force, which is always perpendicular to the object.

Let us now write our equations of motion. Note that since the crate is only moving in the x-direction (horizontal), there is no acceleration in the y-direction (vertical), thus $a_y=0$.

$+\uparrow \sum F_y=0;$

$N_f-490.5+200\sin30^0=0$

$N_f=390.5$ N

For the x-axis, we have the following:

$\rightarrow^+ \sum F_x=ma_x$

$200\cos30^0-0.3(N_f)=(50)(a)$

$200\cos30^0-0.3(390.5)=(50)(a)$

$a=1.12$m/$s^2$

Now that we have the acceleration, we can figure out the velocity and the distance traveled. Note that the acceleration of the crate is constant thus, we can use our kinematics equations.

$v=v_0+at$

(Where $v$ is final velocity, $v_0$ is initial velocity, $a$ is acceleration and $t$ is time)

Substitute the values we know:

$v=v_0+at$

$v=0+1.12(3)$

$v=3.36$ m/s

To find the distance, use the following formula:

$s=s_0+v_0t+\dfrac{1}{2}at^2$

(Where $s$ is final displacement, $s_0$ is initial displacement, $v_0$ is initial velocity, $t$ is time, and $a$ is constant acceleration)

Substitute the values we know:

$s=s_0+v_0t+\dfrac{1}{2}at^2$

$s=0+0(3)+\dfrac{1}{2}(1.12)(3)^2$

$s=5.04$ m

$v=3.36$ m/s
$s=5.04$ m