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

STEP 1

1. The child is sitting 1.40 m from the center of the merry-go-round.
2. The speed of the child is $1.30 \, \mathrm{m/s}$ .
3. The mass of the child is $22.5 \, \mathrm{kg}$ .
4. We are calculating the centripetal acceleration and the net horizontal force exerted on the child.

STEP 2

1. Calculate the centripetal acceleration.
2. Calculate the net horizontal force exerted on the child.

STEP 3

Calculate the centripetal acceleration.
The formula for centripetal acceleration $a_c$ is given by:
$a_c = \frac{v^2}{r}$
where $v = 1.30 \, \mathrm{m/s}$ is the speed of the child and $r = 1.40 \, \mathrm{m}$ is the radius of the circular path.
Substitute the given values into the formula:
$a_c = \frac{(1.30)^2}{1.40}$
Calculate $a_c$ :
$a_c = \frac{1.69}{1.40}$
$a_c = 1.2071 \, \mathrm{m/s^2}$

STEP 4

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 $\boxed{27.66 \, \mathrm{N}}$ .

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