A 51.6 g box is left 17.8 cm from the centre of a merry-go-round. If the box has coefficients of friction of μs = 0.790 and μk = 0.480 with the merry-go-round, what is the maximum speed of the merry-go-round without having the box slide off of the merry-go-round?

To find the maximum speed of the merry-go-round without the box sliding off, we need to consider the maximum static friction force between the box and the merry-go-round.

The formula for static friction is:
Fs ≤ μs * N

Where:
Fs = static friction force
μs = coefficient of static friction
N = normal force

The normal force is the force exerted by the merry-go-round on the box, which is equal in magnitude but opposite in direction to the weight of the box. The normal force can be calculated as:
N = mg

Where:
m = mass of the box
g = acceleration due to gravity (9.8 m/s^2)

Calculating the normal force:
m = 51.6 g = 0.0516 kg
g = 9.8 m/s^2

N = (0.0516 kg) * (9.8 m/s^2)
N = 0.50448 N (rounded to 5 significant figures)

Now we can calculate the maximum static friction force:
Fs ≤ μs * N
Fs ≤ (0.790) * (0.50448 N)
Fs ≤ 0.3981592 N (rounded to 5 significant figures)

Since the static friction force provides the necessary centripetal force to keep the box from sliding off, we can equate it to the centripetal force:

Fs = m * R * ω^2

Where:
m = mass of the box
R = distance from the center of the merry-go-round to the box
ω = angular velocity of the merry-go-round

R = 17.8 cm = 0.178 m

ω = (v / R)
v = maximum speed of the merry-go-round

Substituting the values into the equation:
Fs = m * R * ω^2
0.3981592 N = (0.0516 kg) * (0.178 m) * (v / (0.178 m))^2

Solving for v:
0.3981592 N = 0.0516 kg * v^2
v^2 = (0.3981592 N) / (0.0516 kg)
v^2 = 7.7196147 m^2/s^2
v = √(7.7196147 m^2/s^2)
v ≈ 2.78 m/s

Therefore, the maximum speed of the merry-go-round without having the box slide off is approximately 2.78 m/s.