Newtons Laws EX 6

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 $v_{x_0}=0$ m/s, her velocity at the bottom is $v_x=1$ m/s, the distance she slides is $\mathrm{\Delta }x=3$ m. We calculate the acceleration of the girl down the hill from the equation:

${v_x}^2-{v_{x_o}}^2=2a\mathrm{\Delta }x$

$1^2-0=2\cdot a\cdot 3$

$a=0.17m/s^2$

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

Now, we compute the components of these three forces:

Forces	x-component	y-component
$F_G$	200 cos(-55°) = 115 N	200 sin(-55°) = -163 N
$F_N$	0	$F_N$
$F_f$	$-F_f$	0
$\mathrm{\Sigma }F$	$20\times 0.17=3.4$ N	0

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

$115-F_f=3.4$

$F_f=115-3.4=111.6$ N