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 .

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=

V=

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 =       [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

Cancel the m on both sides;

μs x g =

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 =

v = 11.45m/s

Therefore, the maximum speed he can travel without slipping is 11.45m/s

33: i think it’s b

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