imported>Patrick at 02:44, 15 April 2011

2011-04-15T02:44:02Z

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The girl accelerates down the hill, so we choose the x-direction along the slope of the hill where her initial velocity down the hill is <math>v_{x_0} = 0</math> m/s, her velocity at the bottom is <math>v_x = 1</math> m/s, the distance she slides is <math>\Delta x = 3</math> m. We calculate the acceleration of the girl down the hill from the equation:

<math>{v_{x}}^2 - {v_{x_o}}^2 = 2 a \Delta x</math><br>
<math>1^2 - 0 = 2 \cdot a \cdot 3</math><br>
<math>a = 0.17 m/s^2</math>

We can use Newton's Second Law to find the force of friction <math>F_f</math>. There are three forces acting on the girl: <math>F_N</math>, <math>F_f</math> and <math>F_G</math>. We can draw a free body diagram.

[[image:N_APP_EX_6_SOLN.png|TOP]]<br><br>

Now, we compute the components of these three forces:

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| 200 cos(-55°) = 115 N
| 200 sin(-55°) = -163 N
|-
! <math>F_N</math>
| 0
| <math>F_N</math>
|-
! <math>F_f</math>
| <math>-F_f</math>
| 0
|-
! <math>\Sigma F</math>
| <math>20 \times 0.17 = 3.4</math> N
| 0
|}

Since we are only interested in finding the force of friction, we do not need to consider the y-direction.

<math>115 - F_f = 3.4</math>

<math>F_f = 115 - 3.4 = 111.6</math> N

Newtons Laws EX 6 - Revision history

imported>Patrick at 02:44, 15 April 2011