Question

Sample Problems (Ch.5 Energy) 1A71.4 kg base runner begins his slide into second base when he is moving at a speed of 5-3 m/s. The coefficient of friction between his clothes and Earth is 0.78. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? (b) How far does he slide? 2. A bead of mass m = 8.35 kg is released from point A and slides on the frictionless track shown in Figure. The height of A is ha = 6.20 m. (a) Determine the bead's speed at point B. (b) Determine the network done by the force of gravity in moving the bead from A to C. lo A 1.20m 3. ) m 3. A skier of mass 85 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 83 m up a 42° slope (assumed frictionless) at a constant speed of 5.7 m/s? (b) What power must a motor have to perform this task?

Answer 1

(1.a) Given, The mass of the base runner m = 71.4kg. Initial velocity of the runner is u = 5.3 m/s. Final velocity of the runner is v= 0m/s. As the height of the body remains unchanged the potential energy remains constant. So, the change in kinetic energy is the change in the mechanical energy. The change in mechanical energy is equal to the change in the kinetic energy. The change in kinetic energy is given by. AK = mv - mu? = = m(0m/s)? – 5 mu? =--(71.4 kg)(5.3 m/s)? =-501.4) The loss in mechanical energy is 501.4).

VO The friction force is given by, f = umg . The acceleration is given by, a=-I--umg mm =-ug =-(0.78)(9.81m/s) =-7.65 m/s? The third equation of motion is given by, 12-u? = 2as Here, the displacement is s. s=v-u? 2a _(0m/s)? -(5.3m/s) 2(-7.65m/s) = 0.24 m