An 810 kg race car can drive around an unbanked turn at amaximum speed of 60 m/s

Question

An 810 kg race car can drive around an unbanked turn at amaximum speed of 60 m/s without slipping. The turn has aradius curvature of 150 m. Air flowing over the car's wing exerts adownward-pointing force (called the downforce) of 11500 N on thecar. A)What is the coefficient of static friction between thetrack and the car's tires? B)What would be the maximum speedif no downforce acted on the car?
An 810 kg race car can drive around an unbanked turn at amaximum speed of 60 m/s without slipping. The turn has aradius curvature of 150 m. Air flowing over the car's wing exerts adownward-pointing force (called the downforce) of 11500 N on thecar. A)What is the coefficient of static friction between thetrack and the car's tires? B)What would be the maximum speedif no downforce acted on the car?

Explanation / Answer

According to Newton's second law we can write                      -fs = m (-v2 / R) But fs = sN Therefore we can write                   sN = mv2 / R                   s ( Fd + mg) = mv2 / R Therefore                 s = mv2 / R( Fd +mg)                         = 810 * 60*60 / 150*(11500 + 810*9.8)                         = 1.000 b) If no downward force acting on the car                  s (mg) = mv2 /R                          v = grs                              = 9.8*150*1.00                              = 38.34 m/s

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