The four-wheel-drive all-terrain vehicle has a mass of 320 kg with center of mass G2. The driver has a mass of 82 kg with center of mass G1.

If all four wheels are observed to spin momentarily as the driver attempts to go forward, what is the forward acceleration of the driver and ATV?

The coefficient of friction between the tires and the ground is 0.59.

3.924 m/s²

Explanation:

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - mp. g - mv. g = 0

N1 + N2 = mp. g + mv. g

N1 and N2 are the normal reactions at the wheels, mp is the mass of the driver, mv is the mass of the vehicle and g is the acceleration due to gravity.

Substituting the values in the question,

N1 + N2 = (300+85) x 9.81

N1 + N2 = 3776.85 N

Taking sum of all horizontal forces,

EFx = (mp + mv) x a

μ (N1 + N2) = (mp + mv) x a

μ is the coefficient of friction with a value of 0.4,

Substituting values,

0.4 (N1 + N2) = (300 + 85) a

N1 + N2 = 962.5a

Comparing equations,

962.5a = 3776.85

Therefore the acceleration, a = 3.924 m/s²