Problem Solving Strategies: Mechanical Energy 8.01t Oct 20, 2004Class Problem: Block-Spring System Example 1: A block of mass m is attached to a spring and is free to slide along a horizontal frictionless surface. At t = 0 the block-spring system is stretched an amount x0 > 0 from the equilibrium position and is released from rest. 1. Is the mechanical energy of the block-spring constant? 2. What is the velocity of the block when it first comes back to the equilibrium? 3. What is the period of oscillation of the block?Modeling the Motion: Newton ‘s Second Law • Define system, choose coordinate system • Draw force diagram G Fspring =−kxˆi • Newton’ Second Law 2ˆ:kx dxi −=m dt2PRS Question Which of the following functions x(t) ofthe variable t have a second derivative which is proportional to the negative ofthe function d2x/dt2 ~-x 1. x(t) = (1/2)at2 2. x(t) = Aet/T 3. x(t) = Ae-t/T 4. x(t) = Acos((2π/T)t)PRS Question The first derivative v = dx/dt of the sinusoidal function x(t) = Acos((2π/T)t) is: 1. v(t) = Acos((2π/T)t) 2. v(t) = -Asin((2π/T)t) 3. v(t) = - (2π/T) Asin((2π/T)t) 4. v(t) = (2π/T) Acos((2π/T)t)Modeling the Motion: Newton ‘s Second Law 2 • Equation of Motion: ˆ:kx dxi −=m2dt • Possible Solution: Period of oscillation is T 2π •Position xAcos( t) = T • Initial Position t = 0: A=x0 dx 2π 2π v = =− • Velocity: dt TAsin( Tt) 2π 2π • Velocity at t = T/4 veq=− TA =− Tx0Modeling the Motion: Energy • Choose initial and final states • W =0 nc• Choose zero point for potential energy ( Ux=0) = 0 • Mechanical energy is constant Determine external work ∆+∆ K U0 =Energy Diagram Initial state: at rest with maximum stretch x0 >0 and initial velocity v0 = 0 K = 00 • Kinetic energy U0 = 1kx02 • Potential energy 2 12E0 = kx02 • Mechanical energy 1 Final state: equilibrium position xeq = 0 and velocity veq < 0 K = mv eq eq2 • Kinetic energy • Potential energy Ueq = 1 0 2 • Mechanical energy Eeq = mv eq2 2PRS Question A block of mass m is attached to a spring and is freeto slide along a horizontal frictionless surface. At t = 0the block-spring system is stretched an amount x0 from the equilibrium position and is released from rest.What is the velocity of the block when it first comesback to the equilibrium? 1. Veq = -x0T/4 2. Veq = x0T/4 3. Veq = - (k/m)1/2 x0 4. Veq = (k/m)1/2 x0Mechanical Energy is Constant Eeq −E0 = 1 mveq 2 −1 kx 02 =0 2 2 Solve for velocity at equilibrium position v =− k eq x0m Period T: Condition from Newton’s Second Law 2π− Tx0 = veq Solve for Period = 2π kT = 2π m T m kClass Problem: Block-Spring System with Friction Example 2: A block of mass m slides along a horizontal surface with speed v0. At t = 0 it hits a spring with spring constant k and begins to experience a friction force. The coefficient of friction is variable and is given by µk=bx where b is a constant. Find the loss in mechanical energy when the block has first come momentarily to rest.Experiment 06: Work, Energy and the Harmonic OscillatorExperiment 06: Work, Energy and the Harmonic OscillatorGoals Investigate the work-mechanical energy theorem. Observe how forms of mechanical energy are converted from one to another and lost by non-conservative work. Study the behavior of a simple harmonic motion with a high quality low-loss spring.Equipment setup Use the heavy spring on the force sensor. Put two 250g weights in the cart. Clip motion sensor to other end of track, and support it on a piece of 2x4.Starting DataStudio Create a new experiment. Plug force and motion sensors into the 750 and drag their icons to inputs in the Setup window. Double-click the Force Sensor icon.Force Sensor Set Sample Rate to 500Hz and Sensitivity to Low. Double-click the Motion Sensor Icon.Motion Sensor Ensure to have Acceleration, Position and Velocity checked Set Trigger Rate to 60Hz and calibrate distance to cart when it is resting against the spring. ClickSampling Options No boxes checked! Delayed start on position measurement ! Stop after 10s!Measurement Results Position vs. Time: Measure maximum heights either side of 2nd bounce, calculate loss of potential energy, and friction force. Enter in table! Force vs. Time: Expand force peak around 2nd bounce.Finding Acceleration Up & Down Linear fit to find aup Linear fit to find adownAnalysis Force Peak User-Defined Fit to A*sin(2*pi(x-C)/T)Harmonic Oscillator Unclip motion sensor, raise the force sensor end of track Attach spring to plunger on cart with a binder clip and to the hook on force sensor. Add two 250g weights in the cart. Place motion sensor on table touching other end of track. Set Delayed Start and Auto Stop.Harmonic Oscillator Results Position vs. Time: Measure the period, and calculate spring constant k from M = 0.75 kg. Force vs. Time. Make a plot of force vs. position.Lissajous Patterns Force vs. Position: Find k from a Velocity vs. Position. Linear Fit.Rubber Band Spring - Optional Position vs. Time: Note increased damping. Force vs. Position. Not
View Full Document
Unlocking...