A child sitting 1.50m from the center of a merry-go-around moves with a speed of 1.10m/s .

Part a) Calculate the centripetal acceleration of the child.

Part b) Calculate the net horizontal force exerted on the child ( = 20.5 ).

m=20.5 kg ???

a=v²/R = (1.1)²/1.5 = 0.807 m/s²,
F=m•a = 20.5•0.807=16.5 N

To solve this problem, we will use the formulas for centripetal acceleration and net force.

Part a) The formula for centripetal acceleration (ac) is given by:

ac = v^2 / r

where v is the speed of the child and r is the distance from the center of the merry-go-round.

Given that the speed (v) of the child is 1.10 m/s and the distance (r) from the center is 1.50 m, we can substitute these values into the formula:

ac = (1.10 m/s)^2 / 1.50 m = 0.80 m/s^2

Therefore, the centripetal acceleration of the child is 0.80 m/s^2.

Part b) The formula for net force (F) is given by:

F = m * ac

where m is the mass of the child and ac is the centripetal acceleration.

Given that the net force (F) exerted on the child is 20.5 N, we need to rearrange the formula to solve for the mass (m) of the child:

m = F / ac

Substituting the given values into the formula:

m = 20.5 N / 0.80 m/s^2 = 25.625 kg

Therefore, the mass of the child is approximately 25.625 kg.

To calculate the centripetal acceleration of the child, we can use the formula:

a = v^2 / r

Where:
a = centripetal acceleration
v = tangential velocity
r = radius of the circular motion

Given:
v = 1.10 m/s
r = 1.50 m

Plugging the values into the formula, we get:

a = (1.10 m/s)^2 / 1.50 m

Calculating this, we find that the centripetal acceleration of the child is approximately 0.81 m/s^2.

Now, let's move on to Part b.

The net horizontal force exerted on the child can be calculated using Newton's second law of motion:

Fnet = m * a

Where:
Fnet = net force
m = mass of the child
a = centripetal acceleration

Given:
Fnet = 20.5 N
a = 0.81 m/s^2

To find the mass of the child, we need to rearrange the equation:

m = Fnet / a

Plugging in the values, we get:

m = 20.5 N / 0.81 m/s^2

Calculating this, we find that the mass of the child is approximately 25.31 kg.

Therefore, the net horizontal force exerted on the child is 20.5 N.