2SCALAR PRODUCT or DOT PRODUCT Dot Product is defined as the magnitude of 1st ( ) times scalar component of 2nd vector ( ) along direction of 1st. Essentially, the dot product gives you information about how much of each vector lies along the direction of the other.THIS IS NOT THE SAME AS ADDING 2 VECTORS YIELDING A RESULTANTThe reason the dot product is used in physics is because the operation between certain vector quantities produce meaningful physical answers such as WORK.it’s a scalar result where Ф is the angle between a and b.
3Directional properties of the Dot Product include: Since the dot product involves the cosine thenIF Ф = 90o THEN PRODUCT = 0IF Ф = 0o THEN PRODUCT = MAXIMUMThis is consistent in that when vectors are perpendicular neither lies along the other, therefore an answer of zero results.Directional properties of the Dot Product include:andUnit vector form of dot product:if you distribute, this would reduce to…*Note that there is no direction associated with result but answer can be negative depending on angle.
4Calculate the dot product of the following vectors and find the angle between them: A = -3i + 5j B = 6i +14j
5More Calculus - Derivative of a Product: When taking the derivative of two functions multiplied together, the derivative is: The 1st function times the derivative of the 2nd plus the derivative of the 1st times the 2nd function.Example…find dy/dx*Could have ‘foiled’ and then performed power rule as well
6f (x) = f ( g(x) ) f‘ (x) = f‘( g(x) ) (g'(x)) Chain Rule:The chain rule is used when there is a function within a function.f (x) = f ( g(x) ) f‘ (x) = f‘( g(x) ) (g'(x))Think of the functions f and g as ``layers'' of a problem. Function f is the ``outer layer'' and function g is the ``inner layer.'' Thus, the chain rule tells us to first differentiate the outer layer, leaving the inner layer unchanged (the term f'(g(x))) , then differentiate the inner layer (the term g'(x) ).Derivative of outer times derivative of inner with respect to xThe inner layer, g(x) is (x3 + 5)The outer layer f(x) is (x)2
7Example:Find the derivative of f’function of a function
8Work Done by a Constant Force Work is defined as an external force (F) moving through a displacement (Δr). How much force lies along the movement of an object.
9Positive & Negative Work In all 4 cases, the force has the same magnitude and the displacement of the object is to the right with the same magnitude. Rank the situations from most positive to most negative.
10A crate of mass, M, is dragged along a level rough surface a distance, x, by a force, F as shown. The coefficient of friction is uk.Find the net work done on the crate in terms of given variables and constants.
11Work done by a varying force If F(x) = 4x2 then find the work done on a particle that moves from x = 1m to x = 5m.
12Example:Suppose a mass moves with a trajectory defined bythe position vectorFind the work done by the force,over the interval from t = 1 to t = 2.
14A force, F(x) = 2-4x, acts on a 7. 0kg mass A force, F(x) = 2-4x, acts on a 7.0kg mass. What is the final speed of the mass as it is moved from x=5m to x=2? Assume mass starts from rest at t=0.
15Hooke’s Law Work done by Spring Negative means that the force opposes the displacement from equilibriumIf block is pulled to right, the force by spring is NOT constant via Hooke’s Law. Therefore, the work done by spring must use avg force or be integrated as:Fs
16A plot of spring force vs displacement reveals a slope equal to spring constant, k
17k1k1k2k1k2k2Two springs are attached to a block in series and parallel as shown above. Determine the effective spring constant for each situation in terms of k1 and k2 .What would spring constant be if a mass was attached to a massless spring that stretched a distance x?
18What if same mass was attached to 2 springs as shown What if same mass was attached to 2 springs as shown? How would stretch, x, differ for each spring?
19PowerRate at which work is done or energy is transferred
20A 4kg particle moves along the x-axis A 4kg particle moves along the x-axis. Its position varies with time according to x = t + 2t3, where x is in meters and t is in seconds. Find the the power being delivered to the particle at any time tOR
21If a projectile thrown directly upward reaches a maximum height h and spends a total time in the air T, the average power of the gravitational force during the trajectory is:a) 2mgh / Tb) -2mgh / Tc) 0d) mgh / Te) -mgh / T
23Conservative & Non-conservative Forces The work a conservative force does on an object in moving it from A to B is path independent - it depends only on the end points of the motion.Force of gravity and the spring force are conservative forces. Conservative forces ‘store’ energy…available for kinetic energyThe work done by non-conservative (or dissipative) forces in going from A to B depends on the path taken. Friction is non-conservative. Non-conservative forces don’t ‘store’ energy.
24CONSERVATION OF NRG Total energy in a closed system remains unchanged Work done by NC forces or friction is positive in the above formula. No need to put in minus sign. Work done by friction occurs on left side as minus but becomes + when taken to other side.Energy worksheet
26U(x) vs Restoring Force Relationship As mass oscillates the net force is composed of gravity and the spring force. The net restoring force is shown in the graph.Net force at equilibrium = 0. When slope of U(x) is negative, force is positive and vice versa.
27Potential Energy and Conservative Restoring Forces (gravity, spring) The instantaneous restoring force is equal to the negative derivative of the potential energy function.
28In the case of a mass oscillating on a horizontal frictionless surface we can verify this relationship:
29Example1A certain spring is found not to obey Hooke’s Law; it exerts a restoring force F(x) = -60x-18x2. a) Calculate the potential energy function U(x) for this spring. Let U = 0 when x = 0.b) An object with mass 0.90kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00m to the right to stretch the spring and released. What is the speed of the object when it is 0.50m to the right of x=0?
30Potential energy diagram states of equilibrium: Points of equilibrium are where the force is zero (slope = zero).x3 and x5 are points of stable equilibrium or energy wells. If the system is slightly displaced to either side the forces on either side will return the object back to these positions.x6 is a position of neutral equilibrium. Since there is no net force acting on the object (slope of U(x) = 0) it must either possess only potential energy and be at rest or, it also possesses kinetic energy and must be moving at a constant velocity.x4 is a position of unstable equilibrium. If the object is displaced ever so slightly from this position, the internal forces on either side will act to encourage further displacement instead of returning it back to x4.
31Example2A 5-kg mass moving along the x-axis passes through the origin with an initial velocity of 3m/s. Its potential energy as a function of its position is given in the graph.a) How much total energy does the mass have as it passes through the origin? b) Between 2.5m and 5m, is the mass gaining speed or losing speed?c) How fast is it moving at 7.5m?d) How much potential energy would have to be present for the mass to stop moving?
32Turning pointsPositions where potential energy equals the total mechanical energy, Umax= E, are called turning points
33A particle moves along the x-axis according to the following potential energy function: Find the positions of equilibrium for the particle.