Overview
Chapters 1 through 6 MICRO-management of an object's motion   Newton's Laws \( \sum \vec{F} = m\vec{a} \) plus equations of motion Simulations Computer models Detailed moment-by-moment trajectories Movies/videogames ('frame-by-frame' scene creation) ('simple' only if forces are constant)	Chapters 7 through 9 MACRO-management of an object's motion   Energy approach Relate 'initial' and 'final' conditions directly (can still often use to reverse engineer intermediate results if needed) Can handle non-constant forces Handle abrupt events (explosions, collisions)

 

Forms of Energy

 

Kinetic Energy
Object of some mass m travelling at some speed v Kinetic energy: K = ½ mv2   UNITS Metric (MKS): kg m2 / s2   →   Joules (J) Metric (CGS): g cm2 / s2   →   erg Note that 'g' here represents grams English: calories : 1 c = 4.184 J so 1 J = 0.239005736 c Calories : 1 C = 1000 c = 4184 J ('food calories') BTU : 1 BTU is about 1055 J (about 252 calories) Stands for British Thermal Unit	Examples 100 kg person walking at 2 m/s : K = (0.5)(100)(2)2 = 200 J 2 gram (0.002 kg) bullet at 300 m/s : K = (0.5)(0.002)(300)2 = 180 J 1000 kg vehicle at 60 mph (26.82 m/s) : K = (0.5)(1000)(26.82)2 = 359660 J

 

Other Forms of Energy
Type	Example	Amount
chemical	1 Snicker's candy bar	271 C = 271000 c = 1,130,000 J
chemical	6 inch Subway club sandwich	1,600,000 J = 1.6 MJ
chemical	gasoline	47.2 MJ/kg (132 MJ/gallon)
electrical	supercapacitor	1 MJ/kg
electro-chemical	Lithium-ion battery	1 MJ/kg (phone: ~8000 J)
electro-chemical	lead-acid car battery	0.1 MJ/kg ( ~3 MJ )
nuclear	U-235	79,500,000 MJ/kg
E=mc2	anti-matter	90,000,000,000 MJ/kg

 

The concept of WORK done by a force
A 50 kg crate is being pulled with a force of F=100 N at a 37o angle as shown. If it is initially moving to the right at 1 m/s, how fast will it be moving after displacing 2 m to the right? (Assume no friction here.)
Let's do this completely symbolically and morph the solution into one that focuses on the energy of the crate. Newton's Laws plus equations of motion result in: \( \frac{1}{2}mv^2 = \frac{1}{2}mv^2_o + F_{pull} d \cos{\theta} \) That final term must be the energy that the pulling force added to the object. The energy added (or removed) by a force acting on an object is called the WORK done by that force on the object.

 

Another Case : Object Sliding Down Frictionless Incline
Object initially sliding down ramp at speed vo. How fast will it be moving after travelling a distance d?
Apply Newton's Laws and rearrange into an energy version: Result: \[ K_f = K_o + F_g d \cos{\phi} \] NOTE that \( \phi \) here is the angle between the direction of the force vector and the direction of the displacement vector, so you may have a 'propagate angles' around the figure.

 

Work and the 'Dot' Product
The energy added (or removed) by a force acting on an object is called the WORK done by that force on the object. Mathematically: we're taking two vectors \( \vec{F} \) and \( \vec{d} \) and creating a scalar using a special type of vector multiplication called the dot product (also known as the scalar product) of the two vectors: \[ \vec{F} \cdot \vec{d} = |\vec{F}|~|\vec{d}|~\cos{\phi} = Fd\cos{\phi} \] where:   F is the magnitude of the force \( F = |\vec{F}| \) so is always positive (or zero)   d is the magnitude of the displacement \( d = |\vec{d}| \) so is always positive (or zero)   \( \phi \) is the (interior) angle between the force and displacement vectors
Scalar (dot) product of two vectors \[ \vec{A} \cdot \vec{B} = AB\cos{\theta} \] where A, and B are just the magnitudes of the two vectors, and θ is the (interior) angle between them. NOTE: since the symbol θ is often used to represent the slope of a ramp or some other angle in the figure for a problem and there are usually multiple forces acting, I usually use a different symbol for the dot product: \( \boldsymbol{\phi} \) (the lowercase Greek letter phi)

 

Example

Suppose we have an object moving up an incline, displacing \[ \vec{d} = (3\hat{i} + 2\hat{j})~m \] (using the coordinate system shown in the figure) Multiple forces may be acting on this object, but we want to determine the work that one particular force did: \[ \vec{F} = (-4\hat{i} + 3\hat{j})~N \] Compute the work done on the object by this force as it displaced the given amount. Use the dot-product definition to determine the angle between the force and displacement vectors.

 

Work-Kinetic Energy Theorem

The idea here is that each force (individually) is adding or removing energy from the object so:   For each force acting on the object, compute the work done by that force on the object: \[ W_i = \vec{F}_i \cdot \vec{d} = Fd\cos{\phi_i} \] Adding all those energies to the initial K of the object yields the final K \[ K_{final} = K_{initial} + \sum W_i \] In words: for each force F acting on the object, we compute the work done by that force (i.e. the energy added or removed from the object) by taking the dot product of that force and the displacement (vector) of the object. Adding all those 'works' to the initial kinetic energy yields the 'final' kinetic energy of the object. Notes: \( \phi_i \) is the (interior) angle between a force and the displacement (It will likely be different for each force acting on the object.) the F and d in the final step are the magnitudes of the vectors. Those terms are always positive (or zero) : never negative. The sign of the work comes from taking the cosine of the angle \( \phi_i \) Positive work: adds kinetic energy (speed) to the object Negative work: removes kinetic energy (speed) from the object Zero work: this force doesn't alter the energy (speed) of the object

 

Example
A 50 kg crate is being pulled with a force of F=100 N at a 37o angle as shown. If it is initially moving to the right at 1 m/s, how fast will it be moving after displacing 2 meters to the right? Assume the coefficient of kinetic friction between the crate and the floor is μk = 0.2
\( K_o = ~ ? \) \( W_{F_{p}} = \vec{F}_p \cdot \vec{d} = ~ ? \) \( W_{F_{N}} = \vec{F}_N \cdot \vec{d} = ~ ? \) \( W_{F_{g}} = \vec{F}_g \cdot \vec{d} = ~ ? \) \( W_{F_{fr}} = \vec{F}_{fr} \cdot \vec{d} = ~ ? \) \( K_f = ~ ? \) \( v_f = ~ ? \)