Physics Question

Question

Jenny (mass = 63 kg) is skiing down a 17° slope. A tailwind is assisting her as she travels down the slow with a force of 18 N (parallel to the slope). The coefficient of dynamic friction between Jenny’s skis and the snow is 0.04. Assuming the ski hill is perfectly smooth and the net force in the vertical (y) direction is zero, what is the resultant (net) external force acting on Jenny and what is her acceleration? Wondering the working out for this question?

Answer 1

Sketch a slope and the wind (18N), friction f, weight mg, and normal force F_N. Notice that only the weight vector is not parallel or perpendicular to the slope, so rotate the coordinate system so that x is parallel, then weight should be broken into components. Sum the forces perpendicular to the slope: ΣF_y = F_N - mg cos 17° = 0, and F_N = (63)(9.8)(0.956) = 590 N. Sum parallel forces, with positive being in the forward direction: ΣF_x = 18 + mg sin 17° - f = ma. Friction f = μF_N = (0.04)(590) = 23.6 N. Net force ma = 18 + (63)(9.8)(0.292) - 23.6 = +175 N. The acceleration is then 175/63 = +2.8 m/s² down the slope.