Question(I) A child sitting 1.40 m from the center of a merry-go-round moves with a speed of $1.30 \mathrm{~m} / \mathrm{s}$. Calculate $(a)$ the centripetal acceleration of the child and $(b)$ the net horizontal force exerted on the child (mass $=22.5 \mathrm{~kg}$ ).

Studdy Solution
Calculate the net horizontal force exerted on the child.
The formula for the net horizontal force $F$ is given by:
$$F = m \cdot a_c$$
where $m = 22.5 \, \mathrm{kg}$ is the mass of the child and $a_c = 1.2071 \, \mathrm{m/s^2}$ is the centripetal acceleration calculated previously.
Substitute the given values into the formula:
$$F = 22.5 \times 1.2071$$
Calculate $F$:
$$F = 27.66 \, \mathrm{N}$$
The centripetal acceleration of the child is $\boxed{1.2071 \, \mathrm{m/s^2}}$ and the net horizontal force exerted on the child is $\boxed{27.66 \, \mathrm{N}}$.