11/23/15Oregon State University PH 211, Class #251 What is a restoring force? 1.A binding force that reverses the action of an explosion. 2.A force that opposes the displacement of an object from a point of equilibrium. 3.A force that replaces dissipated energy. 4.The crew that cleans up the stadium after a big football game. 5.None of the above.

11/23/15Oregon State University PH 211, Class #252 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: F = –kx What does this mean? It means that the farther we displace (compress or stretch) the spring from its rest (equilibrium) state, the more force we must exert to do that: Half the distance displaced 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?

11/23/15Oregon State University PH 211, Class #253 A 15-kg. plate is attached to the front of a car with a spring (k = 240 N/m). The car accelerates at a constant 2 m/s 2. How far does the spring compress? cm cm m m 5.None of the above.

11/23/15Oregon State University PH 211, Class #254 Suppose you stretch (or compress) an object on a 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, F max = kA and thus the magnitude of a max : |a max | = F max /m = kA/m

11/23/15Oregon State University PH 211, Class #255 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 can’t use 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, v max ?

11/23/15Oregon State University PH 211, Class #256 Another “Investment-Grade” Force —Another Place to Store Potential Energy The oscillation motion 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). How much energy can we store in such a spring? We know the work done when pushing directly against the constant force of local gravity (W = F·s = mg  y). But unlike local gravity, the force needed for a displacement in spring is a function of the displacement: F = –kx

Spring (or “elastic”) Potential Energy It’s the potential energy associated with position of a stretched or compressed ideal spring: U s = ½k(Δx) 2 Note the SI units will be (N/m)·m 2, or N·m, or J. Note where Δx must be measured from—always from the “rest position” of the spring (i.e. where the spring is neither stretched nor compressed). In the absence of external forces that do work on the system, Conservation of Mechanical Energy can still hold: ΔE mech = Δ K + Δ U g + Δ U s = 0 11/23/157Oregon State University PH 211, Class #25

11/23/15Oregon State University PH 211, Class #258 The fully-developed Work-Energy Theorem looks like this now—three “sub-accounts” in the Mechanical Energy “bank:” E mech = K T + U G + U S = (1/2)mv 2 + mgh + (1/2)kx 2 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): W ext =  E mech Or, arranged differently: E mech.f = E mech.i + W ext Fully detailed: (1/2)mv f 2 + mgh f + (1/2)kx f 2 = (1/2)mv i 2 + mgh i + (1/2)kx i 2 + W ext

11/23/15Oregon State University PH 211, Class #259 A compressed spring (k = 1.00 kN/m) is used to launch a block vertically to a height of 12.0 m above its launch position (the block is not attached to the spring— merely resting on it). The same block is launched again, but this time the spring is compressed half as much as the first time. What height (again, above its launch position) does the block reach this time? m m m m 5.None of the above.