Some sliding rocks approach the base of a hill with a speed of 12 m/s. The hill rises at 36° above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and 0.65, respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

Solution 30E Let us have a look at the following figure. Let the mass of the rocks be m. They are sliding up the hill at an acceleration of a. Since the rocks are sliding up, hence the force of kinetic friction will act downward slope of the hill. If ks the coefficient of kinetic friction, then the frictional force acting on the rocks is F k= mkcos 36 . Moreover, the downward weight on the rocks is mgsin 36 .0 Therefore, the total force acting on the rocks when they are sliding up the hill is F T = mgsin 36 + mgcok 36 0 ma = mgsin 36 + mgcos 36 0 k 0 0 a = gsin 36 + gcks 36 0 0 a = g(sin 36 + cok 36 ) a = 9.80 (0.58 + 0.45 × 0.80) 2 a = 9.21 m/s Therefore, the acceleration of the rocks as they are sliding up the hill is 9.21 m/s . (b) Let us now calculate the downward acceleration of the rocks once they start sliding down. The acceleration in this case will be a = gsin 36 cos 36 0 down k The negative sign before the 2nd expression appears because as the rocks are sliding down, the kinetic friction force will act upward. 0 0 Therefore, a down = 9.80(sin 36 0.45cos 36 ) a down = 9.80(0.58 0.36) a down = 2.15 m/s 2 Therefore, the down acceleration of the rocks is 2.15 m/s . 2 0 Now, component of weight acting downward = mgsin 36 = 0.58mg The force due static friction = sgcos 36 0 = 0.65 × 0.80mg = 0.52mg Since the weight component is more than the static friction force, the rock will slide down the hill.