PES 1110 Fall 2013, Spendier Lecture 19/Page 1

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1 PES 1110 Fall 2013, Spendier Lecture 19/Page 1 Today: - Hooke s Law (chapter section 7.7) - Elastic potential Energy (chapter section 8.4) Demo: The effect of the weight hanged on a coil spring on the extension of the spring A coil spring is hanged with a weight hanger and pointer attached to its end. Different known weights with equal increments (in this case 500g, 1000g, 1500g, 2000g according to the coil's flexibility) are hanged and the readings of the pointer with the respective weights are marked on a sheet of paper. The results show that equal weight increments stretch the spring equally (linear relationship) what we call Hooke's law. Take in account that this linear relationship between weight and length displacement of the spring is limited to the working range of the spring (too small weights will not stretch the spring at all whereas too big weights will compromise its flexibility). Hooke s Law: The force needed to stretch or compress a spring increases linearly with stretching distance Hooke s law uses the stretching distance of the spring. The force increases with stretching distance. So I will be using a slightly different notation as your textbook. Your book calls this distance x, but I will call it s for how far the spring has been stretched. F sp = k s where s = l l 0 k = spring constant, Unit: N/m (lower case k) s = stretching distance (how far has I been stretched) l 0 = equilibrium length (spring length when spring is unstretched) l = length of the spring after force is applied In spring problems put always zero at the position where the spring is unstretched.

2 PES 1110 Fall 2013, Spendier Lecture 19/Page 2 Example 1: A horizontal 50N force is applied to a 100N/m spring whose unstretched length is 0.5m. What is the spring s length after the force has been applied? Example 2: A 5-kg mass is attached, as shown, to a 100N/m spring whose unstretched length is 0.5m. If the mass is pushed 0.3 m to the left from the equilibrium position, what is the magnitude and direction of the force exerted by the spring on the mass? When you stretch a spring, the spring pulls back. When you compress a spring it will push back. So springs can push and pull.

3 PES 1110 Fall 2013, Spendier Lecture 19/Page 3 Work to stretch a spring: The spring force is a variable force so how do we find the work needed to stretch the spring? We need to find the area underneath the force versus position graph. We have an initial stretching distance s 1 and then we will stretch the spring to a distance s 2. The force applied to pull this spring between s 1 and s 2 increased linearly. Hence we plot the equation of a straight line thought the origin F sp = k s To find the work, we find the area. We can subtract the big triangle from the little triangle W s2f2 s1f1 s2ks2 s1ks1 (used Hooke s law: F = k s) W ks2 ks1 The force changes when you compress or stretch a spring. What happens to work? Mathematically the difference is that s is positive when you stretch and s is negative when you compress a spring. What happens when you square this negative number? It becomes positive. So in terms of work stretch and compress is always the same.

4 PES 1110 Fall 2013, Spendier Lecture 19/Page 4 Elastic Potential Energy: Potential energy due to a spring In spring problems put always zero at the position where the spring is unstretched, since they have zero potential when they are unstretched. We pull a box from 0 to s 1 to s 2. There is potential energy here. When I pull the mass and I hold it in place, nothing happens until I let go. The mass will move back and will gain kinetic energy. When I was holding the mass at the rest the kinetic energy was in the spring stored in from of elastic potential energy. You can think of conservative forces. While I am pulling, the spring does work. This work is saved. It does not matter if I wait 5 min or 5 years, the instant I let go then you see this work manifest itself in form of kinetic energy. Just a while ago, we calculated the force needed to stretch the spring. This corresponds to the work I have to do when I pull the spring. But I do not do conservative work, the spring does conservative work. It is the work the spring does on the mass which is saved. The elastic force exerted by the spring is saved or stored - this force points in the negative direction! W s F s F s ks s ks Elastic work converted to potential energy Wel ks2 ks 1 But this is exactly what we need since the work done by a conservative force is: W el = - U el Uel ks elastic potential energy

5 PES 1110 Fall 2013, Spendier Lecture 19/Page 5 Conservation of Elastic Energy: If the spring is the only force which is doing work, then the total energy will not change, it will be conserved. E i = E f 1 1 E K U mv ks el (kinetic energy and elastic potential energy) Then the conservation of energy gives: E i = E f mvi ksi mv f ks f (here the mass will not cancel from every term) Example 3: A 10 kg mass slides across a frictionless, horizontal floor going 5m/s when it collides with a k = 500N/m spring. What is the maximum compression of the spring? Here the mass is slowing down. What is the reason, in form of energy, for the mass to slow down? There is no friction! When it hits the spring its kinetic energy is changed into what from? It is changed into elastic potential energy. There are of course three forces acting on the mass, the normal force, the weight, and the spring (elastic) force. However, I can apply the conservation of energy here since the normal force and the weight do no work in this problem. Both are perpendicular to the motion. To apply the conservation of energy theorem, the spring needs to be the only force which does work. Elastic potential energy bungee jumping: