1.
North^+
East > -
Fnet=Fn-Fg + Ff - Fapp
^
^
0
0
Fnet=Ff
ma=Mmg(cancel out the masses)
a=mg
^
vf*-vi*=2ad
0*- (-25m/s)*=2a(45m)
-625 m*/s*/90m=90m(a)/90m
-6.94 m/s*=a
-6.94m/s*/(-9.81m/s*)=M(-9.81m/s*)/-9.81m/s*
0.71=M
The coefficent of friction between the road and the car's tires is 0.71.
2.
North^+ Convert
velocity from m/s to
East > -
km/h:
50km
(1h) (1000m)= 50,000
Fnet=Ff
h (3600s) (1km) 3600
ma=Mmg(cancel out the masses)
=13.89m/s
a=Mg
a=(0.65)(-9.81m/s*)
a=-6.38m/s*
vf*-vi*=2ad
0*-(-13.89m/s)*=2(-6.38m/s*)d
-192.9321m*/s*/-12.76m/s*=-12.76m/s*(d)/-12.76m/s*
d=15.1m
The distance the car travels is 15.1 metres east.
3.
North ^+
East > +
Fnet=Fn+Fg+Ff+Fapp
^ ^
0
0
Fnet=Fapp
ma=Fapp
vf*-vi*=2ad
0*-(25m/s)*=2(a)(1.5m)
-625m*/s*/3m=3m(a)/3m
-208.3m/s*=a
ma=Fapp
80kg(-208.3m/s*)=-16,666.7 N
The seatbelt provides a 1.7 x10^4 N force backwards.
4.
North ^ -
East > +
vix=5m/s(cos20degrees) viy=5m/s(sin20degrees)
= 4.696 m/s
=1.710m/s
d=1/2at* + vi(t)
1.5m-1.5m=1/2(9.81m/s*)t* + 4.698m/s(t)-1.5m
0=4.905m/s*(t*) + 4.698 m/s(t) -1.5m
~Use the quadratic formula~
t= -4.698+- (squareroot of) [4.698* -(4)(4.905)(-1.5)] /2(4.905)
t= 0.25 s
t= -1.21 s (extraneous root)
v=d/t
d=vt
d=(4.698m/s)(0.25 s)
d=1.17 m
The person flew a horizontal distance of 1 metre.
