A child rolls a 0.600-kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 m/s. When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 m/s. Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

Solution 84P Step 1: Initially, the energy possessed by the ball was purely Kinetic. Once it climbs the ramp, the kinetic energy will be converted to potential energy and some amount of work done due to frictional force. So, we can write, KE initialW fPE top We know that, KE = ½ mv + ½ I 2 initial Where, m - mass of the basket ball v - initial velocity of the ball I - moment of inertia of the ball - Angular velocity of the ball 2 Where, ½ mv = translational kinetic energy 2 ½ I = Rotational kinetic energy of the ball We know that, I = mr for the basketball since, it is considered as a hollow sphere. Angular velocity of the ball, = v/r Where, r - radius of the ball 2 2 2 2 Then, ½ I = ½ × × mr × (v/r) = mv 2 2 2 Therefore, KE initial mv + mv = mv Therefore, mv = mgh - W --------------------- (1) f Where, g - acceleration due to gravity h - maximum vertical height of the ball in the ramp Step 2: Similarly, if the ball climbs down, the PE at the top will be converted to kinetic energy and work done due to friction. So, we can write, PE top= KE final W f Work done due to friction in both cases will be same since, the coefficient of static friction is same for both the upward and downward motion of the ball in the ramp. Therefore, K E final mv’ 2 Where, v’ - final velocity of the ball 2 Therefore, we can write, PE top= mv’ + W f 2 That is, mgh = mv’ + W - f -----------------------(2)