Final answer:
The maximum acceleration the conveyor belt can have without the crate slipping is 4.21 m/s², calculated using the coefficient of static friction and the normal force on the crate. If the acceleration of the belt exceeds this value, the crate will slip, and its acceleration would be determined by other external forces.
Step-by-step explanation:
The question asks for the maximum acceleration of a conveyor belt that a 15 kg crate can tolerate without slipping, given the coefficient of static friction (μs) is 0.43.
To find this, one must use the equation for static friction, which is fs = μsN, where N is the normal force, equivalent to the gravitational force on the crate (mg). With an acceleration of 0, N equals the weight of the crate, which is mg = 15 kg × 9.8 m/s2 = 147 N. The maximum frictional force without slipping is therefore fs(max) = μs × N = 0.43 × 147 N = 63.21 N. Now, fs(max) equals the mass times the maximum acceleration without slipping (amax), so we have amax = fs(max) / m = 63.21 N / 15 kg = 4.21 m/s2.
If the acceleration of the belt exceeds the maximum acceleration, the crate will begin to slip, and the force of friction will now be kinetic rather than static. The kinetic friction force (μkN), however, does not influence the acceleration of the crate itself; the crate would accelerate along with the belt but at a lesser rate, determined by the net external forces acting on it.