Tut Exercise A 5.10kg block is set into motion up an inclined plane with an initial speed of v, = 7.70 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of0 = 30.0 to the horizontal. /\\'/d ,>/ (a) For this motion, determine the change in the block's kinetic energy. (b) For this motion, determine the change in potential energy of the blockEarth system. (c) Determine the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction? Parl1 of 6 - Conceptualize The kinetic energy change is negative and on the order of 100 J. The gravitational potential energy change is positive and smaller in magnitude. We expect a friction force of several newtons and a friction coefcient of perhaps 0.5 to 0.8. Pan 2 of 6 - Categorize There is enough information to find the acceleration of the block, but we do not need to use this formula to solve the problem. Instead, we can think about the conservation of energy as a way of nding the friction force and the friction coefficient. Part 3 of 6 - Analyze We consider the blockplaneplanet system between an initial point just after the block has been given a push and a final point when the block comes to rest. (a) The change in kinetic energy is given by _ 1 z 1 2 =K K- imv imv- AK f I 2 f 2 I =o~%(5.1o / 5.1 kg)( 7.70 .I / 7.7 rn/s)2= -19.6 x ,2 ~151 Your response differs from the correct answer by more than 10%. Double check your calculations. J. (a) The change in kinetic energy is given by AK = KF - Ki = = mvp2 - = mv.2 -0 - 25.10 9 5.1 K9 7.70 2 7.7 m/s = -19.6 x -151 Your response differs from the correct answer by more than 10%. Double check your calculations. J. Part 4 of 6 - Analyze (b) The change in gravitational potential energy is given by AU = mgyf - 0 = 5.1 P 5.1 kg)(9.80 m/s2 ) 3.00 3 m sin 30.00 = 74.97 / P 75 J . Part 5 of 6 - Analyze (c) The nonisolated system energy model can be described using the following equation. AK + AU =_Wother - fid - fid . The force of friction is the only other unknown force, so we find it from fk = -AK - AU J ) - C m and, performing the calculations, we find that fk N . m Submit Skip (you cannot come back). Submit