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) Fin

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)

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Helga 3 years 2021-08-31T00:22:40+00:00 1 Answers 54 views 0

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    2021-08-31T00:23:48+00:00

    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

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