WORK, POWER, MACHINES & ENERGY
Work is the product of the component of the force exerted on an object in the direction of the displacement and the magnitude of the displacement. W = FΔd
WORK In order for work to be done, three things are necessary: There must be an applied force. The force must act through a certain distance, called the displacement. The force must have a component along the displacement.
Read the following statements and determine whether or not they represent examples of work. A teacher applies a force to a wall and becomes exhausted. A book falls off a table and free falls to the ground. NO, displacement doesn’t occur Yes, displacement in the direction of force
A truck carries a box in it’s bed 100 m. NO, This is not an example of work. The force is upward on the box but the displacement is along the ground.
You pull your luggage on a cart that makes an angle of 30º for 5 m Yes, but only in direction of the displacement 5 m So the force that does the work is the component of the force along the ground or Fx So Work = Fd = (Fcos Θ)d
W = F d W = Nm W = (Newtons )(meters) The units of work are; W = Joule (J) In customary; W = foot pounds
Matt lifts a 80 kg barbell upward for 1 meter at a constant speed, how much work does he do? F = w = mg = (80 kg) (10m/s/s) = 800 N What force must Matt provide ? W= Fd = (800 N) (1 m) =800 J
What work is done by a 60 N force in dragging the bag a distance of 50 m when the force is transmitted by a handle making an angle of 30 with the horizontal? F = FcosΘ d = 50m W= FcosΘd W= (60 N)(cos 30º)(50m) W= 2598 J FcosΘ F = 60 N Θ W= F ∙ d
POWER is the rate at which work is done. Power = (work) (time) P = W t P = Joules sec P = J/s = Watts = W In customary; Power = horsepower= hp 760 W = 1 hp
What is the man’s power in lifting a 3.0 kg object through a vertical distance of 1.6 m in 10 sec? F = w = mg d = 1.6 m t = 10 s P = W = Fd t t P = (3 kg) (10 m/s/s) (1.6 m) 10 sec P = 4.8 WAbility to do work?
Energy is the ability to do work or that which can be converted into work.. When something has energy, it is able to perform work or, in a general sense, to change some aspect of the physical world. Pg. 66
PE = m g h In mechanics we are concerned with two kinds of energy : KINETIC ENERGY: KE, energy possessed by a body by virtue of its motion. Units: Joules (J) POTENTIAL ENERGY: PE, energy possessed by a system by virtue of position or condition. KE = ½ mv 2 Units: Joules (J)
Example: Find the kinetic energy of a 3200 N automobile traveling at 20.8 m/s? F g = 3200 N v = 20.8 m/s m = W/g = 320 kg KE = ½ mv 2 = ½ (320 kg) (20.8m/s) 2 = 6.92 x10 4 J
Example: A 250 g object is held 200 mm above a workbench that is 1 m above the floor. Find the potential energy relative to a. the bench top m = 0.25 kg h = 0.2 m PE = mgh = 0.25 kg (10 m/s 2 ) (0.2m) = 0.50 J b. the floor h = 1.2 m PE = mgh = 0.25 kg (10 m/s 2 ) (1.2m) = 3.00 J
Today: Notes: Work and Power Tuesday: Energy Quiz Tomorrow: Notes: Conservation of Energy Monday: Roller Coaster Lab