A 65-kg ice skater coasts with no effort for 75 m until she stops. If the coefficient of kinetic friction between her skates and the ice is , how fast was she moving at the start of her coast?
Solution 83GP: We have to determine the initial velocity of the ice skater as she stops after covering a distance of 75m, due to presence of friction. Step 1 of 5 Concept: Newton’s second law: The net force F acting on an object of mass m produces an acceleration a in that object. Mathematically, F = ma. Frictional force also acts on the surface of the bodies in contact whenever their is relative motion between between the two bodies. Free body diagram of a body gives the diagrammatical representation of all the forces acting on the body in terms of magnitude and diagram, under the given situation. The kinematic equation relating the initial velocity (u)and final velocity (v) of an object, accelerating by an acceleration (a) in time (t) is given as, v = u + at Step 2 of 5 Figure below show the free body diagram of the ice skater. Mass of the ice skater m = 65 kg Weight of the ice skater acting vertically downward due to gravity W = mg = 65g Frictional force acting between the ice skater and surface F fr Normal reaction exerted by the ground on the ice skater F N Distance covered by the ice skater s = 75 m Coefficient of kinetic friction of skates k 0.10 Step 3 of 5 As there are only two forces acting in the vertical direction and both the forces are opposite in direction and also there is no motion in the vertical direction, therefore both the forces would balance each other. F N = mg The magnitude of frictional force would be, F fr = kFN = kg ....(1)