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18 April, 15:32

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 The force is then withdrawn and the body moves a distance of 60m in the next 5 seconds.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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Answers (2)
  1. 18 April, 15:51
    0
    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. 18 April, 18:11
    0
    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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