Work and Energy Unit 4
W = Fcosq Dx Lesson 1 : Work Done by a Constant Force When a force acts on an object while displacement occurs, the force has done work on the object. The magnitude of work (W) is the product of the amount of the force applied along the direction of displacement and the magnitude of the displacement.
N . m = Joule (J) + Work - Work If the force has a component in the direction of the displacement. If the force has a component in the opposite direction of the displacement. Units of Work Determining the Sign of Work
Rank the following situations in order of the work done by the force on the object, from most positive to most negative. [Displacement is to the right and of the same magnitude.] Example 1
4.0 kg 5.0 m m = 0.30 30o Find the work done by all forces as a 4.0 kg mass slides 5.0 m down a 30o incline where the coefficient of kinetic friction is 0.30. Example 2
F (N) x (m) Work done is the area under the graph Graphical Analysis of Work F WF = FDx Dx
Since W = Fcosq Dx and A . B = ABcosq then W = F .Dx Work is a Scalar (Dot) Product
A particle moving in the xy plane undergoes a displacement Dx = (2.0 i + 3.0 j) m as a constant force F = (5.0 i + 2.0 j) N acts on the particle. ^ ^ ^ ^ Example 3 a) Calculate the magnitudes of the displacement and the force. b) Calculate the work done by force F.
xf W = SFxDx xi Lesson 2 : Work Done by a Varying Force
xf ò lim S FxDx = Fxdx Dx 0 xi xf ò W = Fxdx xi As Dx approaches 0, Therefore,
A force acting on a particle varies with x, as shown above. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. Example 1
The interplanetary probe shown above is attracted to the Sun by a force given by 1.3 x 1022 F = - x2 Example 2
This equation is in SI units, where x is the Sun-probe separation distance. Determine how much work is done by the Sun on the probe as the probe-Sun separation changes from 1.5 x 1011 m to 2.3 x 1011 m.
~ 60 squares Graphical Solution Each square = (0.05 N)(0.1 x 1011 m) = 5 x 108 J
Hooke’s Law force exerted by spring Fs = -kx position relative to equilibrium position spring constant in N/m Work Done by a Spring Negative sign signifies that the force exerted by spring is always directed opposite to the displacement.
stretched spring equilibrium position compressed spring
xf ò (-kx)dx = Ws = ½ kx2 ò Ws = Fsdx = xi ½ kx2 Work done by the spring force is positive because the force is in the same direction as displacement.
xf ò (-kx)dx = ½ kxi2 - ½ kxf2 Ws = xi xf xf ò ò Ws = Fappdx = kxdx = ½ kxf2 - ½ kxi2 xi xi Generalized Work Done by Spring Generalized Work Done on Spring
Example 3 A common technique used to measure the spring constant (k) is shown above. The spring is hung vertically, and an object of mass m is attached to its lower end. Under the action of the “load” mg, the spring stretches a distance d from its equilibrium position.
a) If a spring is stretched 2.0 cm by a suspended mass of 0.55 kg, what is the spring constant of the spring ? b) How much work is done by the spring as it stretches through this distance ? c) Suppose the measurement is made on an elevator with an upward vertical acceleration a. Will the unaware experimenter arrive at the same value of the spring constant ?
Example 4 If it takes 4.00 J of work to stretch a Hooke’s Law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm.
Example 5 A light spring with spring constant 1200 N/m is hung from an elevated support. From its lower end a second light spring is hung, which has spring constant 1800 N/m. An object of mass 1.50 kg is hung at rest from the lower end of the second spring. a) Find the total extension distance of the pair of springs.
b) Find the effective spring constant of the pair of springs as a system. We describe these springs as in series.
Example 6 ^ ^ A force F = (4xi + 3yj) N acts on an object as the object moves in the x-direction from the origin to x = 5.00 m. Find the work W = F . dx done on the object by the force. ò
Work done by SF is xf ò SW = SF dx xi Lesson 3 : Work-Kinetic Energy Theorem
(by chain-rule) dx dv dv vf xf xf xf ò ò ò ò dt dx dt SW = SW = SW = SW = m dx ma dx mv dv m dx SW = ½ mvf2 – ½ mvi2 vi xi xi xi
KE = ½ mv2 SW = KEf – KEi = DKE Kinetic Energy Work - Kinetic Energy Theorem If work done on a system only changes its speed, the work done by the net force equals the change in KE of the system.
Example 1 A 6.0 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.
Example 2 A man wishes to load a refrigerator onto a truck using a ramp. He claims that less work would be required to load the truck if the length L of the ramp were increased. Is his statement valid ?
Example 3 A 4.00 kg particle is subject to a total force that varies with position as shown above. The particle starts from rest at x = 0. What is its speed at a) x = 5.00 m b) x = 10.00 m c) x = 15.00 m
vf - vi ax = t Lesson 4 : Situations Involving Kinetic Friction SFx = max (SFx)Dx = (max)Dx Dx = ½ (vi + vf) t
( vf - vi ) (SFx)Dx = m ½ (vi + vf) t t This is not work because Dx is displacement of a particle – the book is not a particle ! -fkDx = DKE (SFx)Dx = ½ mvf2 – ½ mvi2 (SFx)Dx = -fkDx = ½ mvf2 – ½ mvi2 = DKE
DKE = -fkd + SWother forces d = length of any path followed -fkd = DKE KEf = KEi - fkd + SWother forces DKE in General OR
Example 1 A 6.0 kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 12 N. a) Find the speed of the block after it has moved 3.0 m if the surfaces in contact have a coefficient of kinetic friction of 0.15.
b) Suppose the force F is applied at and angle q as shown below. At what angle should the force be applied to achieve the largest possible speed after the block has moved 3.0 m to the right ?
DEint = fkd Change in Internal Energy due to Friction The result of a friction force is to transform KE into internal energy, and the increase in internal energy is equal is equal to the decrease in KE. DEsystem = DKE + DEint = 0 -fkd + DEint = 0
Example 2 A 40.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find a) the work done by the applied force b) the increase in internal energy in the box-floor system due to friction
c) the work done by the normal force d) the work done by the gravitational force
e) the change in kinetic energy of the box f) the final speed of the box.
Lesson 5 : Power Same amount of work done Time interval is different
W P = Dt W dW P = lim = Dt 0 Dt dt dW F . dx F . v P = = = dt dt Average Power time rate of energy transfer Instantaneous Power
1 W = 1 J/s = 1 kg . m2/s3 1 hp = 746 W Units of Power SI unit of power is J/s or the Watt (W). In the U.S. customary system, the unit of power is the horsepower (hp).
1 kWh = (103 W)(3600 s) = 3.60 x 106 J The kilowatt-hour (kWh) The energy transferred in 1 h at the constant rate of 1kW = 1000 J/s. * Note that a kWh is a unit of energy, not power.
An elevator car has a mass of 1600 kg and is carrying passengers having a combined mass of 200 kg. A constant friction force of 4000 N retards its motion upward, as shown above. Example 1
a) What power delivered by the motor is required to lift the elevator car at a constant speed of 3.00 m/s ? b) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.00 m/s2 ?
Example 2 Find the instantaneous power delivered by gravity to a 4 kg mass 2 s after it has fallen from rest.
Example 3 Find the instantaneous power delivered by the net force at t = 2 s to a 0.5 kg mass moving in one dimension according to x(t) = 1/3 t3.
Consider a car of mass m that is accelerating up a hill, as shown above. An automotive engineer measures the magnitude of the total resistive force to be ft = (218 + 0.70v2) N where v is in m/s. Determine the power the engine must deliver to the wheels as a function of speed. Example 4
100 kg The 100 kg box shown above is being pulled along the x-axis by a student. The box slides across a rough surface, and its position x varies with time t according to the equation x = 0.5t3 + 2t, where x is in meters and t is in seconds. Example 5 : AP 2003 #1 a) Determine the speed of the box at time t = 0.