A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionles

Question

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.55 m/s to the right relative to the plank.

Required:
What is the velocity of the plank relative to the surface of the ice?

in progress 0
Thu Hương 5 years 2021-08-30T03:23:30+00:00 1 Answers 23 views 0

Answers ( )

  1. thugiang
    0
    2021-08-30T03:24:50+00:00
    Reply

    Answer:

    The speed of the plank relative to the ice is:

    v_{p}=-0.33\: m/s

    Explanation:

    Here we can use momentum conservation. Do not forget it is relative to the ice.

    m_{g}v_{g}+m_{p}v_{p}=0 (1)

    Where:

    • m(g) is the mass of the girl
    • m(p) is the mass of the plank
    • v(g) is the speed of the girl
    • v(p) is the speed of the plank

    Now, as we have relative velocities, we have:

    v_{g/b}=v_{g}-v_{p}=1.55 \: m/s (2)

    v(g/b) is the speed of the girl relative to the plank

    Solving the system of equations (1) and (2)

    45v_{g}+168v_{p}=0

    v_{g}-v_{p}=1.55

    v_{p}=-0.33\: m/s

    I hope it helps you!      

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )