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7 June, 21:36

You put a 3 kg block in the box, so the total mass is now 8 kg, and you launch this heavier box with an initial speed of 4 m/s. How long does it take to stop

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Answers (2)
  1. 8 June, 00:28
    0
    The box stops at zero speed.

    Final Velocity = 0, Initial speed (s) = - 4 m/s

    Therefore = change in velocity = Vf - Vi. (0 m/s - 4 m/s) = - 4 m/s

    Change in velocity = - 0.4 m/s

    Gravity g = 9.8 m/s^2

    Mass = 0.8 g

    -4 ms divided by 9.8 ms^2 * (0.8) = 0.51 s

    It takes 0.51 seconds to stop
  2. 8 June, 00:36
    0
    0.7 secs

    Explanation:

    In this question, the speed does not change as the mass changes. So we can use

    Δt=Δ∨x/χgμ ... equ 1

    To stop, the final speed will be 0

    Therefore,

    Δvx=vf-vt

    Δvx=0-4m/s

    = - 4m/s

    Now substitute the various values in equ 1

    Δt=Δ∨x/χgμ

    Δt = - 4m/s / (9.8m/s∧2) (0.6)

    Δt=0.7 secs
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