PHY111/211 Pre Laboratory: Conservation of Mechanical Energy Objectives: Investigate the work-energy relation and the conservation of mechanical energy Activities: Review work, potential energy, work-energy and the conservation of mechanical energy, then answer the problems that follow. Work: The work done by a force on a particle whose displacement is Ax is represented by the area under the force-versus-position curve. If the force is not constant (variable) the work done by the force is W =F(Ax) or W = [ F, dx = area under the force-versus-position graph Example: Use basic mathematics or calculus to find the work (Newton-meters - N-m or Joules - J) done by the force shown. F (N) Basic: W = Area = height x width = (10N)(5m) = 50 N-m = 50J 10 Calculus: W =[ F - dx = [ION - dx = 10NX= =50N -m-ON -m=50J x (m) 0 5 Example: Use basic mathematics or calculus to find the work (Newton-meters - Nom or Joules - J) done by the force shown. F (N) Basic: W = Area = 1/2 height x width = 12 (10N)(5m) = 25 N-m = 25J 10 Calculus: first find F(x) function. F(x) = mx+b=2x(N)+0 x (m) 5 W =[F . dx = [2x(N -m)-dx =1/2(2x7)& =25Nom-ON.m=25N.m Gravitational Potential Energy and Work: Here is the relation between the work done by gravity and gravitational potential energy of an object moving down a frictionless surface. The force of gravity, w = mg, and the vertical displacement, S (yr - yi), are in the same direction; thus, the force of gravity does positive work on the object. Warm = F . Ay = (mg)-(As cose) my However, the potential energy of the object decreases as it moves down the surface toward ground. yF St APE = -W = mg(y, -y,) NOTE: yr-y is negative in this case Conversely, if the object is pulled up the surface by an external force, the potential energy of the object increases; the pulling force does positive work and the force of gravity does negative work.10-2 Work-Kinetic Energy Theorem: The work done on an object by a net force equals the change in kinetic energy of the object W NET = A(KE) = - mv,2 - -mv Example: A mother pulls her child and sled (10kg) from rest with a force of 15N for a distance of 10m. (a) How much work does the mother do? (b) How fast is the child moving after 10m? (a) WNET = F .5 =15N-10m =150N . m =150Joules (b) W NET = A(KE) = = mv- -mv; 150N . m =- (10kg),2 -0 10m (2)150N - m) =5.47m/s = 11mph 10kg Conservation of Mechanical Energy. Let us consider the work done on a skier skiing down a hill. The positive work done by the force of gravity (+War) equals the decrease in the potential energy of the skier (-APE) and equals the increase in the kinetic energy of the skier (+AKE). W NET =-APE) =+A(KE) Thus, as the potential energy of the skier decreases, her kinetic energy increases as she speeds toward the bottom of the hill. The sum of the kinetic energy and the potential energy of the system is called the total mechanical energy EMEC. PE + KE, = PE, + KE, = const. For our conservative system (frictionless ski slope) the total mechanical energy remains constant during the down hill motion of the skier. This is the law of conservation of energy. Example: A motor pulls a 10kg block of ice up a 20" frictionless incline for a distance Ax = 20m. Find (a) the force required to pull the block up the plane at a constant velocity, (b) the work done by the motor, (c) the increase in potential energy of the block of ice and (d) the work of gravity. (a) EF, =0= mgsin 0-F =0= F = mgsin 0=33.5N motor (b) W. = Fmatar ' Ax = (33.5N)(20m) =670N -m =670J Ax O (c) APENdoc = mgh = mg(Arsin 0) = (10kg)(9.8m/s')(20msin 209) h = Ax sing my sini APE back = 670J Alternate method: APE Nock = W.=670J (d) Wa=-APE=-670J wemyName: Sec: PHY111/211 Pre Laboratory: Conservation of Mechanical Energy Problem #1: (a) Calculate how much work is done by the force of gravity when a skier moves up a 30 frictionless incline if the skier weighs 680 N (150lbs) and the length of the incline is 10m. (b) Explain why this work is negative. Hint: Find the component of gravitational force that is parallel and down the incline. Note that the component of force due to gravity is down the plane. 10m Displacement Force of gravity 5m down the plane Problem #2: (a) What is the change in potential energy of the 680 N skier as she is pulled from the base to the top of a 5m high hill. (b) How does your answer compare to Problem #1? Explain why. (c) Why is the potential energy positive? Use words and concepts to explain, not the equation W = -AU Problem #3: (a) Find the final velocity of a 680 N skier at the base of a frictionless hill if she starts out from rest at a point 5m above the base. (b) Repeat Part (a) for a 900 N skier. (c) Is the kinetic energy at the base of the hill the same for each skier? Explain. Selected answers: 1. -3400J 2. +3400J 3. (a) 9.9 m/s