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What is a restoring force? A. A binding force that reverses the action of an explosion.

What is a restoring force? A. A binding force that reverses the action of an explosion. B. A force that opposes the displacement of an object from a point of equilibrium. C. A force that replaces dissipated energy. D. The crew that cleans up the stadium after a big

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What is a restoring force? A. A binding force that reverses the action of an explosion.

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  1. What is a restoring force? A. A binding force that reverses the action of an explosion. B. A force that opposes the displacement of an object from a point of equilibrium. C. A force that replaces dissipated energy. D. The crew that cleans up the stadium after a big football game. E. None of the above. OSU PH 211, Before Class #25

  2. What is a restoring force? A. A binding force that reverses the action of an explosion. B.A force that opposes the displacement of an object from a point of equilibrium. C. A force that replaces dissipated energy. D. The crew that cleans up the stadium after a big football game. E. None of the above. OSU PH 211, Before Class #25

  3. One form of ideal spring provides a linear restoring force (a vector) directed opposite to the spring’s displacement relative to its equilibrium point. In other words: Fspring.mass = –kxmass What does this mean? It means that the farther we displace (compress or stretch) the spring from its rest (equilibrium) state—where we define x = 0—the more force we must exert to do that: Half the displacement needs half the force. Five times the displacement needs five times the force, etc. The spring constant, k, is characteristic of the spring itself— measuring its strength or “stiffness.” What are the units of k? OSU PH 211, Before Class #25

  4. A 15-kg. plate is attached to the front of a car with a spring (k = 240 N/m). The car is accelerating at a constant 2 m/s2. How far does the spring compress while the car is accelerating? A. 12.5 cm B. 13.3 cm C. 1.25 m D. 1.33 m E. None of the above. OSU PH 211, Before Class #25

  5. A 15-kg. plate is attached to the front of a car with a spring (k = 240 N/m). The car is accelerating at a constant 2 m/s2. How far does the spring compress while the car is accelerating? A.12.5 cm B. 13.3 cm C. 1.25 m D. 1.33 m E. None of the above. OSU PH 211, Before Class #25

  6. Suppose you displace a mass on a spring—either stretching or compressing the spring—then release it and observe its motion. Q: Where is the magnitude of its acceleration a maximum —and where is it a minimum? A: The maximum acceleration of a mass oscillating on a spring occurs at the extremities of its motion (i.e. where x = ±A), because that’s where the force is a maximum. And knowing the spring constant, k, we can easily compute the magnitude of that force, Fmax = kA and thus the magnitude of amax: |amax | = Fmax/m = kA/m OSU PH 211, Before Class #25

  7. Q: Where is the oscillating mass’s speed a maximum —and where is it a minimum? The force on—and therefore the acceleration of—the mass varies at every different point in the spring’s oscillation. We canNOT use constant-acceleration kinematics to analyze the motion of the mass. A: The maximum speed of the mass is at the equilibrium point. That is, v is maximum (and F and a are zero) where x is zero. But how can we compute that maximum speed, vmax? Via an analysis of mechanical energy.… OSU PH 211, Before Class #25

  8. Another Way to Store Potential Energy The oscillation of a mass on an ideal spring continues undiminished if there is no friction—or other external force—doing work on the mass. That is, mechanical energy is conserved in an ideal spring. We get that energy back—by releasing the spring (just as we get the work back from gravity, by releasing an object to fall). So, how much energy can we store in such a spring? We know the work done when pushing directly upward, against the constant force of local gravity (W = –FG·s = –|mg||Dy|cosf = mgDy). But unlike local gravity, the force exerted by a spring to oppose displacement of a mass is a function of that displacement: Fspring.mass = –kxmass So, to find the work invested in the displacement, we must integrate: W = – ∫Fspring.mass.xdx = – ∫–kxmassdx = (1/2)kx2 OSU PH 211, Before Class #25

  9. Spring (or “elastic”) Potential Energy • This is the elastic potential energy associated with the displacement of a stretched or compressed ideal spring: • Uelas = US = ½k(x)2 • Note the SI units will be (N/m)·m2, or N·m, or J. • Note where x must be measured from—always from the “rest position” of the spring (i.e. where the spring is neither stretched nor compressed). That position is always defined as x = 0. • In the absence of external forces that do work on the system, Conservation of Mechanical Energy can still hold: • ΔEmech = ΔK + ΔUg+ ΔUs = 0 OSU PH 211, Before Class #25

  10. So the fully-developed Work-Energy Theorem now looks like this—now with three “sub-accounts” in the Mechanical Energy “bank:” • Emech = KT + UG + US • = (1/2)mv2 + mgh + (1/2)kx2 • Any change in the total bank balance is due to “deposits” and/or “withdrawals”—work done by external forces (forces other than gravity or ideal springs): Wext = Emech • Or, arranged differently: Emech.f = Emech.i + Wext • Fully detailed: (1/2)mvf2 + mghf + (1/2)kxf2 • = (1/2)mvi2 + mghi + (1/2)kxi2 + Wext OSU PH 211, Before Class #25

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