A body at rest is acted upon by a force for 20 seconds. The force is then withdrawn and the body moves a distance of 60m in the next 5 secon

Question

A body at rest is acted upon by a force for 20 seconds. The force is then withdrawn and the body moves a distance of 60m in the next 5 seconds. If the mass of the body is 10 kg, calculate the magnitude of the force.​

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Mít Mít 3 years 2021-08-30T15:52:30+00:00 2 Answers 12 views 0

Answers ( )

  1. giabao
    0
    2021-08-30T15:53:37+00:00
    Reply

    Answer:

    6 N

    Explanation:

    Given:

    v₀ = 0 m/s

    v = 60 m / 5 s = 12 m/s

    t = 20 s

    Find: a

    a = Δv / Δt

    a = (12 m/s − 0 m/s) / 20 s

    a = 0.6 m/s²

    F = ma

    F = (10 kg) (0.6 m/s²)

    F = 6 N

  2. Jezebel
    0
    2021-08-30T15:53:40+00:00
    Reply

    Answer:

    6N

    Explanation:

    From the principle of conservation of momentum.

    The momentum initiated by the Force F in 20sec is the impulse and is given as;

    F×t= F×20 = 20F

    Similarly the mass 10kg acquires a momentum of m×v; where m is mass =10kg and v is the velocity of the body.

    The velocity of the body, V = Distance/ time =60/5=12m/s

    Hence the momentum is;

    10×12= 120Ns

    By conservation of momentum;

    20F =120

    F = 120/20 = 6N

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