Question

A 41.0-kg child swings in a swing supported by two chains, each 3.04 m long. The tension in each chain at the lowest point is 358 N. (a) Find the child's speed at the lowest point. m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.) N (upward)

Answer 1

Answer:

V = 4.826m/s, 716N

Explanation:

At the lowest swinging point, the net force acting on the child is equal to the centripetal force and it could be represented as

F = mv^2/r

2T-mg =mv^2/r

r(2T-mg) = mv^2

Divide both sides by m

r(2T-mg)/m = mv^2/m

r(2T/m-g) = v^2

V= √ r(2T/m-g)

Where v is the velocity

r is the length of the chain

m is the mass of the child in kg

T is the tension in Newton

g is the acceleration due to gravity

Given that g = 9.8m/s^2

T = 358N

m = 41.0kg

r = 3.04m

Substituting the values into the formula

V = √ 3.04( 2*358/41 -9.89

V = √ 3.04 ( 716/41 - 9.8 )

V = √3.04 ( 17.463-9.8 )

V = √3.04( 7.6634)

V = √23.2967

V = 4.826m/s

For the second part which is the tension in the two chains

N - m*g = m*(v^2 / r)

N - (41)*(9.81) = (41)*(4.826^2 / 3.04)

N - 402.21 = 41×7.66

N - 402.21 = 314.112

N = 402.21 + 314.112

N = 716.332 newton

Approximately = 716N

Or alternatively, since there are two chains holding the swing, of which each chain is acted upon by a 358N tension. Hence = 2T

2*358 = 716N