Download Andrew Rosen Work: Product of force and displacement (scalar) W

Andrew Rosen
Work: Product of force and displacement (scalar)
W = Fd
-
“Work for the joules”
Work is only done in the direction of the force!!!
If something is not moving, then there is no work being done (even if there is a force!)
Area under a F vs. D graph gets you work
If there is a diagonal force, you must break it into components (Fdcos )
The force for a person running upstairs, for example, is their weight
Power: Rate at which work is done
P=
=
“Power is in Watt?!”
Energy: The ability to do work (measured in joules)
Kinetic Energy: Energy of motion
KE =
-
Mass and KE = Direct
Velocity and KE = Square relationship
Must be positive
Work-Energy Theorem: The net work done on an object is equal to its change in KE
-
Don’t try to shortcut it! Find KE before as well as after. Then find the change.
W=
W=
Potential Energy: Stored Energy
Gravitational Potential Energy:
PE = mgh
-
h can be used as a reference point... unless you hate yourself
It can be positive or negative based upon the reference point
Law of Conservation of Energy: Within a closed, isolated system, energy can change form but the toal
amount of energy is constant. Energy cannot be created or destroyed, but it can change forms.
-
“What is it that I love about constants?! Oh, THEY’RE CONSTANT!”
The total kinetic and potential energy in a system is known as the total mechanical energy
Andrew Rosen
Springs and Other Fun Stuff:
Hooke’s Law: F = kx
-
-
k is the spring constant and x is the spring length from the equilibrium position
In a F vs. x graph, k can be found by taking the slope
A high k value is a stiff spring while a low k value is a weak spring
The area under the curve = work done
o (½)(F)(x) = W
o W = (½)(kx)x  W = (½)(kx2)
The F is typically the weight for a mass hung from a vertical spring
The work done in compressing or extending a spring = the energy stored in the spring
PEs = ½kx2
-
PE at the top is equal to the KE at the bottom of a pendulum
Work = PEs
Work done by Friction: Wfr = Wfd = Q (heat)
-
If a problem states that 800J should be produced and mathematically, only 600J are produced,
200J must be lost to friction
Conservation of Mechanical Energy:
Etotal = KE1 + PE1 = KE2 + PE2 (+Q if there is friction)
(Extremely Shiny) Examples:
1) You hold a penguin by exerting a force of 6.02x1023N on the penguin upwards. How much work
has been done? How much work has been done if you decide to produce the same upwards
force but at the same time travel east 400m at a constant .00001 m/s?
Answer: NO WORK! It’s not in the direction of the force!
2) Andrew decides to chuck a (very light) elephant at a certain someone. Since Andrew does not
wish to waste his strength, he puts the elephant on a golden-plated spring (of doom). If the
spring constant is 200 N/m, the elephant’s mass is 0.20kg, and the distance spring compresses is
.30m, find the kinetic energy as the elephant is released from the spring into the air. Also find
the (initial) velocity of said elephant as it is launched.
Answer: PEs = (1/2)(200 N/m)(.30m)2 = 9J  KE = 9J
9J = (1/2)(.20kg)(v)2
V = 9.5 m/s
*Explanation: The energy stored as PE is converted entirely to KE as the elephant is launched
3) Find the sum of two and two
Answer: 4
Andrew Rosen
4) In a trebuchet, a counterweight is dropped on a lever to launch a projectile at a target. If the
projectile weighs 80N, the counterweight falls 17m, the average drag force on the projectile is
2N, the target is 75m away, and the velocity needed for the projectile to go through the target is
15m/s, what is the mass of the counterweight?
Answer: mg = 80N, h = 17m, m = ?, Ff = 2N, d = 75m, v = 15m/s
PEi = KEf + Q
(m)(9.8 m/s2)(17m) = (1/2)(
)(15m/s)2 + (2N)(75m)
m = 6.4kg
*Explanation: The potential energy of the falling counterweight was converted into the kinetic
energy of the projectile while the remaining energy was lost to friction
5) You are in your car when suddenly a rollercoaster track comes out of nowhere! Luckily,
someone put a spring at the end so you can survive. Assuming no friction or air resistance or
anything else you can think of, find:
a. The velocity (V1) given the information in the diagram
Etotal = KEi + PEi = KEf + PEf
(200kg)(9.8 m/s2)(30m) + (1/2)(200kg)(2m/s)2 = (1/2)(200kg)(v2) + (200kg)(9.8 m/s2)(20m)
V = 14.14 m/s
b. The spring constant given the information in the diagram
(200kg)(9.8 m/s2)(30m) + (1/2)(200kg)(2m/s) = (1/2)(k)(8m)2
k = 1850 N/m (theoretically speaking, this makes no sense, but go with the flow)
c. The energy lost to friction at V2 if you ignore my previous statement and assume there is
work done by friction going on
(200kg)(9.8 m/s2)(30m)+(1/2)(200kg)(2m/s)=(1/2)(200kg)(18m/s)2+(200kg)(9.8m/s2)(5m)+Q
Q = 17000J
Andrew Rosen
6) A box slowly falls down a cliff onto a straight path. This hill was 5m high. The initial velocity of
the box was 0 m/s, and its mass was 2kg. Assuming the hill had no friction, and the straight path
had a coefficient of friction equal to .3, find the distance the box travels when it comes to a halt
(without using kinematic equations).
Answer: (2kg)(9.8m/s2)(5m) = 98J = PEi = Etotal
Ff =
= (.3)(2kg*9.8m/s2) = 5.88N
= Wf  Ffd =
 98J = (5.88N)(d)
d = 16.67m
7) Repeat problem number 6 with kinematic equations
Answer: (2kg)(9.8m/s2)(5m) = 98J = PEi = Etotal
Ff =
= (.3)(2kg*9.8m/s2) = 5.88N
KE = (1/2)(m)(v2)  98J = (1/2)(2kg)(v2)
V = 9.9m/s
F = ma  -5.88N = (2kg)(a)
a = -2.94m/s2
Vf2 = Vi2 + 2ad
(0m/s)2 = (9.9m/s)2+(2)(-2.94m/s2)(d)
d = 16.67m