A car starts rolling down a 1 -in-4 hill (1 -in-4 means that for each 4 m traveled along the sloping road, the elevation change is 1 m). How fast is it going when it reaches the bottom after traveling 55 m? (a) Ignore friction. (h) Assume an effective coefficient of friction equal to 0.10.
Solution 71GP: We have to determine the velocity of the car as it is reaches the of the hill in presence and absence of friction. Step 1 of 8 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. 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), having its displacement (s) is given as, v = u + 2as Step 2 of 8 FIgure below shows the free body diagram of the moving car. m Mass of the car mg Weight of the car acting vertically downward due to gravity Angle made by the inclined surface with the horizontal F frFrictional force acting between the car and surface N Normal reaction exerted by the ground on the car s Distance covered while reaching the bottom = 55 m Step 3 of 8 For simplification of calculation resolving the weight into two components, mgsin Component of weight acting along the plane mgcos Component of weight acting perpendicular to the plane Also, distance covered while reaching the bottom s = 55 m For each 4.0 m horizontal distance covered by the car along the hill, the elevation in the hill is 1.0 m. Therefore, the angle made by the surface with the horizontal is 1 sin = 4 1 1 = sin 4 o = 14.5 o Angle made by the inclined surface with the horizontal is = 14.5 . Step 4 of 8 A] Assuming the friction is absent If friction is not present the only component of weight responsible for accelerating the car down the hill would be mgsin.Using Newton’s second law, we get, ma = mg sin a = g sin a = 9.8 ×sin 14.5 o a = 9.8 ×sin 14.5 o a = 2.45 m/s 2 Acceleration of the car down the hill, in absence of friction, is a = 2.45 m/s2