Sect 8 4 Solving Problems with Energy Conservation

Sect. 8 -4: Solving Problems with Energy Conservation

Conservation of Mechanical Energy K + U = 0 (Conservative forces ONLY!!) or E = K + U = Constant • For gravitational U: Ugrav = mgy E = K 1 + U 1 = K 2+ U 2 (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 y 1 = Initial height, y 2 = Final height, v 1 = Initial velocity v 2 = Final velocity

Conservation of Mechanical Energy K + U = 0 (Conservative forces ONLY!!) or E = K + U = Constant • For elastic (Spring) U: Uelastic = (½)kx 2 K 1 + U 1 = K 2+ U 2 (½)m(v 1)2 + (½)k(x 1)2 = (½)m(v 2)2 +(½)k(x 2)2 x 1 = Initial compressed (or stretched) length x 2 = Final compressed (or stretched) length v 1 = Initial velocity, v 2 = Final velocity

Example 8 -6: Pole Vault Estimate the kinetic energy & the speed needed for a 70 kg pole vaulter to just pass over a bar 5. 0 m high. Assume that the vaulter’s center of mass is initially 0. 90 m off the ground & that he reaches his maximum height at the level of the bar.

Example 8 -7: Toy Dart Gun For elastic forces, Mechanical energy Conservation gives: A dart of mass m = 0. 1 kg is pressed against the spring of a toy dart gun. The spring (spring constant k = 250 N/m) is compressed x = 6. 0 cm & released. The dart detaches from the spring when the spring reaches its natural length (x = 0). Calculate the speed of the dart there. x 1 = 0. 06 m, v 1 = 0, m = 0. 1 kg k = 250 N/m, x 2 = 0, v 2 = ? Get: v 2 = 3 m/s

Example 8 -8: 2 Kinds of Potential Energy A ball, of mass m = 2. 6 kg, s s starts from rest, falls a distance e h = 55. 0 cm before striking a a vertical coiled spring, which it I I compresses an amount l Y = 15. 0 cm. v 1 = 0 v 2 = ? C a Calculate the spring constant k. a The spring has negligible mass. a Ignore air resistance. Measure all a a distances from the point where the a a ball first touches the uncompressed a a spring (take y = 0 at that point). v 3 = 0

v 1 = 0 m = 2. 6 kg, h = 0. 55 m Y = 0. 15 m, k = ? A two step problem: STEP 1: (a) (b) v 2 = ? (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 v 1 = 0, y 1 = h = 0. 55 m, y 2 = 0 Get: v 2 = 3. 28 m/s STEP 2: (b) (c) (both gravity & spring U) v 3 = 0 (½)m(v 2)2 + (½)k(y 2)2 + mgy 2 = (½)m(v 3)2 + (½)k(y 3)2 + mgy 3 = Y = 0. 15 m, y 2 = 0 (½)m(v 2)2 = (½)k. Y 2 - mg. Y Solve for k & get k = 1590 N/m ALTERNATE SOLUTION: (a) (c) skipping (b)

Example: Bungee Jump m = 75 kg, k = 50 N/m, y 2 = 0 v 1 = 0, v 2 = 0, y 1 = h = 15 m + y y = ? Conservation of Mechanical Energy with both gravity & spring (elastic) U (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 +(½)k( y)2 = 0 + mg(15 + y) = 0 + (½)k( y)2 A quadratic Equation for y: Solve & get y = 40 m & -11 m (throw away the negative value) Solve by Fig. (a) to Fig. (c) directly!! Δy + 15 m

Example 8 -9: A swinging pendulum The simple pendulum consists of a small bob of mass m suspended by a massless cord of length ℓ. The bob is released (without a push) at t = 0, where the cord makes an angle θ = θ 0 to the vertical. a. Describe the motion of the bob in terms of kinetic energy & potential energy. b. Calculate the speed of the bob as a function of position θ as it swings back & forth. c. Calculate the speed at the lowest point of the swing. d. Calculate the tension in the cord, FT.

Example – A Spring Loaded Cork Gun • • • A ball of mass m = 35 g = 0. 035 kg in a popgun is shot straight up with a spring of constant k. The spring is compressed y. A = - 0. 12 m, below it’s relaxed level, y. B = 0. The ball gets to maximum height y. C = 20. 0 m above the relaxed end of the spring. (A) If there is no friction, find the spring constant k. (B) Find speed of ball at point B. It starts from rest & speeds up as the spring pushes against it. As it leaves the gun, gravity slows it down. Only conservative forces are acting, so we use Conservation of Mechanical Energy The initial kinetic energy K = 0. Choose both the gravitational potential energy Ug = 0 & elastic potential energy Ue= 0 where the ball leaves the gun. At it’s maximum height, K = 0 again. Note that Two types of potential energy are needed!

• For the ball’s entire trip, Mechanical Energy is Conserved!! or: KA + UA = KB + UB = KC + UC. At each point, U = Ug + Ue so, KA+ Ug. A+ Ue. A = KB+ Ug. B + Ue. B = KC + Ug. C + Ue. C (A) To find the spring constant k, use: KA+ Ug. A+ Ue. A = KC + Ug. C + Ue. C or, 0 + mgy. A + (½)k(y. A)2 = 0 + mgy. C + 0, giving k = [2 mg(y. C – y. A)/(y. A)2] = 958 N/m (B) To find the ball’s speed at point B, use: KA+ Ug. A+ Ue. A = KB + Ug. B + Ue. B or, (½)m(v. B)2 + 0 = 0 + mgy. A + (½)k(y. A)2 , giving (v. B)2 = [k(y. A)2/m] + 2 gy. A; or, (v. B) = 19. 8 m/s

Other forms of energy; Energy Conservation Nonconservative, or dissipative, forces: Friction, Heat, Electrical energy, Chemical energy & more do not conserve mechanical energy. However, when these forces are taken into account, the total energy is still conserved:

Sect. 8 -5: The Law of Conservation of Energy is one of the most important principles in physics. The total energy is neither decreased nor increased in any process. Energy can be transformed from one form to another & from one body to another, but the total amount remains constant. Law of Conservation of Energy • Again: Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only! This is a general Law!!

Sect. 8 -6: Problems with Friction • We had, in general: WNC = K + U WNC = Work done by non-conservative forces K = Change in Kinetic Energy U = Change in Potential Energy (conservative forces only!) • Friction is a non-conservative force! So, if friction is present, we have (WNC Wfr) Wfr = Work done by friction In moving through a distance d, the force of friction Ffr does work Wfr = - Ffrd

When friction is present, we have: Wfr = -Ffrd = K + U = K 2 – K 1 + U 2 – U 1 – Also now, K + U Constant! – Instead, K 1 + U 1+ Wfr = K 2+ U 2 OR: K 1 + U 1 - Ffrd = K 2+ U 2 • For gravitational U: (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 + Ffrd • For elastic or spring U: (½)m(v 1)2 + (½)k(x 1)2 = (½)m(v 2)2 + (½)k(x 2)2 + Ffrd

Example 8 -10: Roller Coaster with Friction A roller-coaster car, mass m = 1000 kg, reaches a vertical height of only y = 25 m on the second hill before coming to a momentary stop. It travels a total distance d = 400 m. Calculate the work done by friction (the thermal energy produced) & estimate the average friction force (assume it is roughly constant) on the car. m = 1000 kg, d = 400 m, y 1 = 40 m, y 2= 25 m, v 1= y 2 = 0, Ffr = ? (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 + Ffrd Ffr= 370 N

Example 8 -11: Friction with a Spring A mass m slides along a rough horizontal surface at a speed v 0. It strikes a massless spring head-on & compresses it a maximum distance X. The spring constant is k. Calculate the coefficient of kinetic friction μk between the block surface.

Example – Block Pulled on a Rough Surface • A block, mass m = 6 kg, is pulled by constant horizontal force F = 12 N. over a rough horizontal surface. Kinetic friction coefficient μk = 0. 15. Moves a distance Δx = 3 m. Find the final speed.

Example – A Block – Spring System • A mass m = 1. 6 kg, is attached to ideal spring of constant k = 1, 000 N/m. Spring is compressed x = - 2. 0 cm = - 2 10 -2 m & is released from rest. (A) Find the speed at x = 0 if there is no friction. (B) Find the speed at x = 0 if there is a constant friction force Fk = 4 N.

Example – A Crate Sliding Down a Ramp • A crate, mass m = 3. 0 kg, starts from rest at height y 0 = 0. 5 m & slides down a ramp of length d = 1. 0 m & incline angle θ = 30° under a constant friction force Fk = 5 N. It continues to move on the horizontal surface afterwards. (A) Find the speed at the bottom. (B) Assuming the same friction force, find the distance on the horizontal a surface that the crate moves after a it leaves the ramp.

Example: A Block-Spring Collision • Block, mass m = 0. 8 kg, is given an initial velocity v. A = 1. 2 m/s to the right. It collides with spring with constant k = 50 N/m. (A) If there is no friction, find the maximum compression distance xmax of spring after the collision. (B) Suppose there is a constant friction force Fk between the block & the surface. The coefficient of friction is μk = 0. 5. Find the maximum compression distance x. C now.

Example: Connected Blocks in Motion • Two blocks, masses m 1 & m 2, are connected by spring of constant k. m 1 moves on a horizontal surface with friction. The system is released from rest, when the spring is relaxed and m 2 is at height h above the floor. The motion eventually stops when m 2 is on floor. Calculate the kinetic friction coefficient μk between m 1 & table.