Question

I need help with problem 39 parts b,c,d

Unit 7. Work and Energy In Class Packet Name: 36. A 2.45 kg rock is dropped from the top of a 15.5 m vertical cliff a. What is the potential energy of the rock relative to the base of the cliff before it is dropped? b. What is the kinetic energy of the rock just before it hits the ground below? c. What is the velocity of the rock at a point 5.00 m above the base of the cliff? d. How much total energy does the rock have half way down? 37. A roller coaster is at the top of a 75.0 m hill. It rolls down the hill on very low friction wheels and climbs up to the top of a 45.0 m hill. Find: (a) the speed of the thing at the bottom of the first hill (b) its speed at the top of the second hill 8. A roller coaster starts at some height that you do not know. It goes down this hill and then goes up a second hill that is 28.5 m above the first drop at a speed of 22.5 m/s. So how high was the initial hill? 39. A roller coaster sits at the top of a hill and is preparing to enter a loop-the-loop as shown to the right. The hill has a height H, the loop has a radius of 0.4H and the roller coaster has a mass of M a. What is the gravitational potential energy of the roller coaster while at the top of the hill? b. What will be the velocity of the roller coaster when it reaches point C? C. What will be the velocity of the roller coaster when it reaches point A? d. What will be the velocity of the roller coaster when it reaches point B? 14

Answer 1

(b) Conserving energy at the top and point 'c'. rightwardsdoublearrow MgH = 1/2 mv^2 Potential energy at the top Therefore v = squareroot 2g H (c) Velocity of roller coaster will be same at the bottom of loop-loop and at 'c' as there is no friction. By this logic Conserving energy at 'c' and at 'A'. 1/2 m (squareroot 2g H)^2 = 1/2 m (v_A)^2 + m_g (2 times R) Energy at 'A' rightwardsdoublearrow 1/2 2g H = 1/2 v_A^2 + g times 2 times 0.4 H Therefore v_A = squareroot (g H.2/2 - 0.8 g H) times 2 v_A = squareroot 2 g H - 1.6 g H v_A = squareroot 0.4 g H \r\n d) Conserving energy at c and B. A, v_A = squareroot 0.4 g H v_C = squareroot 2 g H 1/2 m times 2 g H = 1/2 m v_B^2 + 0.4 mg H v_B = squareroot (m_g H - 0.4 m_g H)/2 v_B = squareroot 1.2 g H