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26 December, 09:05

In the winter sport of curling, two teams alternate sliding 20 kg stones on an icy surface in an attempt to end up with the stone closest to the center of a target painted on the ice. During one turn, a player releases a stone that travels 27.9 m before coming to rest. The friction force acting on the stone is 2.0 N. What was the speed of the stone when the player released it?

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  1. 26 December, 09:26
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    u = 2.36 m/s

    Explanation:

    Parameters given:

    Mass of stone = 20kg

    Distance moved before coming to rest = 27.9m

    Frictional force = 2.0N

    We can first find the acceleration of the stone by using the equation for frictional force:

    F = - ma

    Where m = mass of the object

    a = acceleration

    The negative sign denotes that the Frictional force acts opposite the motion of the stone.

    => 2.0 = - 20*a

    a = - 2.0/20

    a = - 0.1m/s²

    Using one of the equations of motion, we have that:

    v² = u² + 2as

    Where v = final velocity

    u = initial velocity

    s = distance moved.

    In this case, v = 0m/s because the object comes to rest.

    => 0 = u² + (2 * - 0.1 * 27.9)

    0 = u² - 5.58

    u² = 5.58

    u = √ (5.58)

    u = 2.36 m/s

    The initial velocity of the stone is 2.36m/s
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