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 $\Delta x=3$ m. We calculate the acceleration of the girl down the hill from the equation:

${v_{x}}^{2}-{v_{x_{o}}}^{2}=2a\Delta x$

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

$a=0.17m/s^{2}$

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
$\Sigma F$	$20\times 0.17=3.4$ 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