View source for Newtons Laws EX 31

There are four forces acting on this block: <math>F_G</math>, <math>F_N</math>, <math>F_f</math>, <math>F</math>. The block slides up the incline and therefore the force of kinetic friction points down the incline (because friction opposes motion).<br><br>
[[image:N_APP_EX_31_SOLNa.png|TOP]]<br><br>

We select the x-axis parallel to the incline and draw a free body diagram:<br><br>
[[image:N_APP_EX_31_SOLNb.png|TOP]]<br><br>

The components of the forces exerted on the mass m = 5 kg are:

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| 50 cos(-127°) = -30 N
| 50 sin(-127°) = -40 N
|-
! <math>F_N</math>
| 0
| <math>F_N</math>
|-
! <math>F_f</math>
| <math>-F_f</math>
| 0
|-
! <math>F</math>
| 25 cos(-37°) = 20 N
| 25 sin(-37°) = -15 N
|}
<br>
Note that we used <math>g = 10 m/s^2</math> in our computations.
<br><br>
First we have to deal with the y-component. Writing

 <math>-40 - 15 + F_N = 0</math>

we see that <math>F_N = 55 N</math>. Substituting this value for <math>F_N</math> in the equation for the force of kinetic friction we get:

<math>F_{f_{max}} = \mu_K F_N = (0.1)(55) = 5.5 N</math>
<br><br>
Now, looking at the x-component, we write

 <math>-30 + 20 - 5.5 = 5 a</math>

Solving for <math>a</math>, we find <math>a = -3.1 m/s^2</math>.
<br><br>
To find the distance travelled in 2 s, we will use the equations of kinematics. The initial velocity is 6 m/s. We use the following equation for displacement:

 <math>\Delta x = v_i t + \frac{1}{2} a t^2</math>

Substituting, we find:

<math>\Delta x = (6)(2) + \frac{1}{2}(-3.1)(2)^2 = 5.8 m</math>