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physics

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Hockey Puck A 110 g hockey puck sent sliding over ice is stopped in 13 m by the frictional force on it from the ice.
(a) If its initial speed is 5.0 m/s, what is the magnitude of the frictional force?
1 N
(b) What is the coefficient of friction between the puck and the ice?
2

  • physics -

    Vf^1=Vi^2 + 2ad but a= force/mass, so

    you can solve for force.

    coefficent offriction:
    frictionforce=mu*mg solve for mu.

  • physics -

    you have to give formulas not answers

  • Physics -

    v = u + a*t
    so t = (v-u)/a
    x = t (v+u)/2
    substituting for t, x = (v+u)(v-u)/2a
    put in known values:
    13 = (5)(-5)/2a
    a = -25/26 m/s^2

    f = m*a
    u(k) * f(normal) = m*a
    u(k) *m*g = m*a
    u(k) = a/g
    u(k) = -25/26 /9.81 = 0.098
    where u(k) = coefficient of friction

    for magnitude of friction force:
    u(k)*m*g = 0.098 * 0.110kg * 9.81 m/s^2 = 0.106N

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