distance it travels would be 2.86 times its distance of travel on earth. (assume that $mu$ remains the same…

distance it travels would be 2.86 times its distance of travel on earth. (assume that $mu$ remains the same as in the problem.) practice it use the worked example above to help you solve this problem. the hockey puck in the figure, struck by a hockey stick, is given an initial speed of 22.0 m/s on a frozen pond. the puck remains on the ice and slides 1.10 x 10² m, slowing down steadily until it comes to rest. determine the coefficient of kinetic friction between the puck and the ice. exercise hints: getting started | im stuck! an experimental rocket plane lands and skids on a dry lake bed. if its traveling at 79.0 m/s when it touches down, how far does it slide before coming to rest? assume the coefficient of kinetic friction between the skids and the lake bed is 0.710. resources read it

distance it travels would be 2.86 times its distance of travel on earth. (assume that $mu$ remains the same as in the problem.) practice it use the worked example above to help you solve this problem. the hockey puck in the figure, struck by a hockey stick, is given an initial speed of 22.0 m/s on a frozen pond. the puck remains on the ice and slides 1.10 x 10² m, slowing down steadily until it comes to rest. determine the coefficient of kinetic friction between the puck and the ice. exercise hints: getting started | im stuck! an experimental rocket plane lands and skids on a dry lake bed. if its traveling at 79.0 m/s when it touches down, how far does it slide before coming to rest? assume the coefficient of kinetic friction between the skids and the lake bed is 0.710. resources read it

Answer

  1. For the "PRACTICE IT" problem (finding the coefficient of kinetic - friction for the hockey puck):
    • Explanation:
      • Step 1: Identify the kinematic equation
        • We know the initial speed (v_0 = 22.0\ m/s), the final speed (v = 0\ m/s), and the displacement (x - x_0=1.10\times 10^{2}\ m). The kinematic equation (v^{2}=v_{0}^{2}+2a(x - x_0)) can be used to find the acceleration (a).
        • Rearranging the equation for (a), we get (a=\frac{v^{2}-v_{0}^{2}}{2(x - x_0)}). Substituting (v = 0), (v_0 = 22.0\ m/s), and (x - x_0 = 110\ m), we have (a=\frac{0-(22.0)^{2}}{2\times110}).
        • (a=\frac{- 484}{220}=- 2.2\ m/s^{2}).
      • Step 2: Relate the acceleration to the frictional force
        • The frictional force (F_f=\mu_k N), and from Newton's second - law (F = ma), where (F=-F_f) (negative because it opposes motion) and (N = mg) (on a horizontal surface). So, (-\mu_kmg=ma).
        • Canceling out the mass (m) on both sides, we get (\mu_k=-\frac{a}{g}).
        • Substituting (a=-2.2\ m/s^{2}) and (g = 9.8\ m/s^{2}), we have (\mu_k=\frac{2.2}{9.8}\approx0.224).
    • Answer: (0.224)
  2. For the "EXERCISE" problem (finding the sliding distance of the rocket plane):
    • Explanation:
      • Step 1: Find the acceleration due to friction
        • The frictional force (F_f=\mu_k N), and from Newton's second - law (F = ma). On a horizontal surface (N = mg), so (F_f=\mu_kmg) and (a=-\frac{F_f}{m}=-\mu_kg).
        • Given (\mu_k = 0.710) and (g = 9.8\ m/s^{2}), then (a=-0.710\times9.8=-6.958\ m/s^{2}).
      • Step 2: Use the kinematic equation to find the distance
        • We know (v_0 = 79.0\ m/s), (v = 0\ m/s), and (a=-6.958\ m/s^{2}). Using the kinematic equation (v^{2}=v_{0}^{2}+2a(x - x_0)), and solving for (x - x_0) (the distance (d) the plane slides).
        • Rearranging the equation gives (x - x_0=\frac{v^{2}-v_{0}^{2}}{2a}).
        • Substituting (v = 0), (v_0 = 79.0\ m/s), and (a=-6.958\ m/s^{2}), we have (x - x_0=\frac{0-(79.0)^{2}}{2\times(-6.958)}=\frac{-6241}{-13.916}\approx449\ m).
    • Answer: (449)