Strategy The forces acting on the snowboarder are her weight and the contact force of the slope, which has a component normal to the incline and a component along the incline (force of kinetic friction). Because she moves along the slope, the most convenient reference frame for analyzing her motion is one with the x-axis along and the y-axis perpendicular to the incline. In this frame, both the normal and the frictional forces lie along coordinate axes, the components of the weight are mg sin 0 along the slope and mg cos 0 at right angles into the slope, and the only acceleration is along the x-axis (@y = 0). Solution We can now apply Newton's second law to the snowboarder: SF: = max > Fy = may N - mgcos0 = m(0). mg sin 0 - 14\ N = ma From the second equation, N = mg cos 0. Upon substituting this into the first equation, we find Significance Q3 = 9 (sin 0 - 14) cos 0) = 9 (sin 13* - 0.20 cos 13°)= 0.29 m/s2. Notice from this equation that if 0 is small enough or /4k is large enough, a= is negative, that is, the snowboarder slows down. R 6_1 2/2 points (graded) The snowboarder above is now moving down a hill with incline 10.0°. What is the snowboarder's acceleration? Give your answer in m/s2 to two significant figures without units. -0.23 -0.23 What can you see about the snowboarder's motion? She is moving at constant speed. She is speeding up. She is slowing down.