$(1])7$ of $8$

rorce$-$Acceleration

Writing the relationship at $$

$60$ :

${\displaystyle \underset{?}{\overset{?}{}}}{F}_t=$Wcos$60=m{a}_{t2}$

$=>60(9\mathrm{.}81)cos60=60a_{t2}$

$=>a_{t2}=4\mathrm{.}905\frac{m}{s^2}$

$a_{n2}=\frac{v_2^2}{}=\frac{(8\mathrm{.}24{)}^2}{4}=16\mathrm{.}99\frac{m}{s^2}$

$a_2=\sqrt[\phantom{2}]{a_{t2}^2+a_{n2}^2}=17\mathrm{.}69\frac{m}{s^2}$

$(4)$ Normal to the Body

Normal reaction the track

exerts on the skateboarder

at $=60$ :

${\displaystyle \underset{?}{\overset{?}{}}}{F}_n=N-$Wsin$60=m{a}_{n2}$

$N-60(9\mathrm{.}81)sin60=60(16\mathrm{.}99)$

$=>,N=1529N$

$(u_1-u_2^3{)}_p=-F_{f}S=-0\mathrm{.}05N(4\mathrm{.}18)$

SOLVE THIS PROBLEM BUT DO NOT IGNORE FRICTION IT HAS FRICTION $0\mathrm{.}05$