Answer:
F.net/F.frictional= 585.32 N and V=11.43 m/s
Explanation:
First of all finding Frictional force between wheels and surface:
F.net= μs.m.g
As m=85 kg
Then F.net= .69 X 85 X 9.98= 585.32 N
Now finding velocity of skate boarder on this radius:
V= [tex]\sqrt{coefficient of static frition.g.r}[/tex]
V=[tex]\sqrt{0.69 X 9.98 X 19}[/tex]
where r= 19 m
V=11.438 m/s
A skateboarder is attempting to make a circular arc of radius r = 19 m in a parking lot. The total mass of the skateboard and skateboarder is m = 85 kg. The coefficient of static friction between the surface of the parking lot and the wheels of the skateboard is μs = 0.69. What is the maximum velocity, in m/s, he can travel at through the arc without slipping?
11.45m/s
For the skateboarder to travel without slipping there has to exist a centripetal force (F) to keep him on track.
The centripetal force (F) is related to the mass (m) and the acceleration (a) of the skateboarder as follows;
F = m x a ----------------(i)
Where;
a = [tex]\frac{v^{2} }{r}[/tex] [v is the linear velocity and r is the radius of the motion path]
And in this case, the centripetal force is actually the force of friction and is given by;
F = μs x m x g [μs is the coefficient of static friction, g = acceleration due to gravity]
Now substitute F and a into equation(i) as follows;
μs x m x g = m x [tex]\frac{v^{2} }{r}[/tex]
Cancel the m on both sides;
μs x g = [tex]\frac{v^{2} }{r}[/tex]
Make v the subject of the formula;
v = √(μs x g x r) --------------------(ii)
Where;
μs = 0.69
r = 19m
Now, take g = 10m/s² and substitute these values into equation (ii)
v = √(0.69 x 10 x 19)
v = [tex]\sqrt{131.1}[/tex]
v = 11.45m/s
Therefore, the maximum speed he can travel without slipping is 11.45m/s
