Phys 201 1nd Edition Lecture 14 Outline of Last Lecture I. Review for Exam 2Outline of Current Lecture II. Conservation of EnergyIII. Non-Constant ForcesIV. Spring ForceA. Work as Area Under the CurveB. External WorkCurrent LectureConservation of energy:This means that the total energy is constant, even though direction and force are changing. Projectile motion demonstrates conservation of energy. When an object moves in a projectile motion, even thoughit slows down as it rises and speeds as it falls, the energy of every point along the projectile path will be the same. Non-constant forces:Some forces will change direction or magnitude depending on position, even if the motion of the object it is acting on continues in the same direction. For example, if you’re pulling a string attached to a large mass towards you, the closer the object gets to you, the higher you will have to raise your arms to keep pulling it and the harder you will have to pull. Even though the object continues to move in the same direction across the floor towards you, you still have to change the direction and magnitude of the force you are applying on the object to pull it. This means that you are applying a non-constant force.For non-constant forces, you need to consider forces in pieces. This means that the total work done on the object is the sum of the work done at each given point of the motion. Equationally, this looks like; W=ΔW1 +ΔW2…..ΔWn W=F1X1Cos(θ1) + F2X2Cos(θ2)….FnXnCos(θn)Force to stretch a spring:These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Springs are the most common example of non-constant forces. When you pull a spring, the force has to increase as distance the spring is stretched is increased. This means that the force is proportional to the distance that the spring is stretched. The equation for spring force then is;F=kXFor this equation, k is a constant. This means that for one spring, no matter what distance the spring is stretched(X), k will always be the same. The value of k depends on the kind of spring you have. Because kis a factor of force and distance, the units of k are N/m.Calculating average force: To calculate the work done by the force of a spring, we need to consider the change in force. Because force changes at a constant proportion to the distance, we can calculate the work done by the force of the spring by measuring the area under the graph of applied force vs. distance. The graph below shows this measurement. kXXBecause the area that work represents is a triangle, we can calculate work with the area equation for anygiven right triangle.Work=area of triangle  Area of a triangle=.5area of rectangle  Area of rectangle=X*kxWork =.5X*kx =.5kx2External Work:The work used to actually stretch a spring is called external work, and is always equal to the potential energy stored in a spring. Therefore, we know that;Wext= PEstored 0.5kx2=mghThis means that even though the force to stretch a spring is not a constant force, conservation of energy works in springs too because you get back the energy expended on stretching a spring. In example, if a Workmass is attached to a vertical spring and causes the spring to stretch, the potential energy of the spring will increase and gravitational potential energy of mass will decrease.Example: A 1kg object is resting on a 1cm compressed spring on a ramp with an angle of 30◦. When the spring is released, how far up the ramp will the mass go? For this problem we need to consider motion in both the X and Y directions. First we need to list the values that we know in relation to our force equation. Then we plug in our values into our conservation of energy equation and solve. However, because we are using a spring, we have to consider both the gravitational potential energy of the mass and the potential energy of the spring. This means that our conservation of energy equation is now KEi + PEgi + PEsi = KEf + PEgf + PEsfF=kXk=F/X = mgCos(θ)/Xk= (1)(9.81)Cos(30)/.01=849N/mX=.01mX2=.0001KEi + PEgi + PEsi = KEf + PEgf + PEsf.5kX2 + 0 + 0 = 0 + mgyf + 0.5(849)(.0001)=mgyfYf=(.042)/(1*9.81)=.0043M=