PHY 121 (Mechanical Physics) Exam #2 Study Guide
Topics covered in the exam:
1. Newton´s laws with friction
2. Work and energy (nonconservative forces + mechanical energy) 3. Conservation of energy
4. Conservation of momentum and collisions
Friction ⃗f Friction is velocity, surface area independent
∙ Static there is no relative motion between the surfaces of contact; friction is static ( fs )
Always equal to the amount of force applied
Model: fs≤ fsmax=¿ fs≤ μs N
We also discuss several other topics like What is the meaning of p. chytridiomycota?
μs → Coefficient of static friction. Directed opposite to motion.
∙ Kinetic There is nonzero relative velocity between the surface of contact; then friction is kinetic ( fk )
Model: fk=μk N
μk = Coefficient of kinetic friction. Constant dependent on materials. Directed opposite to relative motion.
Work and Energy ∙ Problems based on Newton´s second law: ⃗F =m ⃗a .
Don't forget about the age old question of Why do we care about measurements?
∙ Recall: ⃗a=d ⃗vdt=d2⃗v
∙ System a collection of objects under study whose physics is of interest. ∙ Environment is everything else.
∙ Given ⃗F the equation needs to be solved for ⃗r (position) and ⃗v
1. ⃗F is constant (i.e. weight of an object near Earth, ⃗w=mg ) 2. ⃗F=⃗F(t) is a function of time (i.e. bat hitting a ball, ⃗F=F∗t2) 3. ⃗F=⃗F(⃗r) is a function of position (i.e. gravity, electromagnetic, etc.; long range) ∙ Key relationships to keep in mind: Don't forget about the age old question of What is the purpose of maxwell’s equations?
Don't forget about the age old question of Is bayreuth worth visiting?
a dt=v0+(Fm )t → x (t )=x0+∫0tvdt=x0+v0t+12 (Fm )t2 (for constant
mv dv=12m v2−12m v02 → K=12m v2 If you want to learn more check out What is the meaning of experimental design?
Kinetic energy (scalar quantity): ∫ v0
Work: ∫x 0
WorkEnergy theorem: W=ΔK
W=−ΔU (where U is potential energy)
2nd WorkEnergy theorem: ΔK =−ΔU → ΔK+ΔU =0 . Thus, total mechanical energy is defined as E=K+U . Then, ΔE=0 when energy is conserved. Potential energy a conservative force where: F=−dU We also discuss several other topics like What is the meaning of violent motion?
dx (in 1 dimension).
Also, W=∫ x 0
F(x )dx=∫ x0
dxdx=−(U( x f)−U ( x0) )=¿W =−ΔU (system doing
work on its environment)
∙ Potential energy is associated with a system, not an isolated object. PE is a form of energy that is “retrievable” in the sense that PE stored in a system can be converted into work (KE). PE is the work gravity does: Ug( y )=mgy.
∙ Elastic energy (Hook´s law) The force of an elastic body undergoing a linear displacement Δ ⃗x is opposite to the displacement and proportional to its magnitude on its environment: ⃗F=−K Δ ⃗x where K is a constant
Stretch: ⃗F=−K (x−x0)^x
Compress: ⃗F=K(x 0−x )^x
Elastic potential energy: Ue( x)=12K ( x−x0 )2
∙ Adding nonconservative forces:
W=W c+W nc→−ΔU+W nc=ΔK → ΔK+ΔU=W nc ; where ΔK+ΔU=ΔE → ΔE=Wnc
Thus, the energy equation says:
KF +UF=K0+U 0+Wnc
∙ The momentum of a particle of mass m with velocity ⃗v is ⃗P=m ⃗v ∙ In Newton´s second law language: ⃗F=d ⃗P
When m is constant d ⃗P
dt=md ⃗vdt=m ⃗a→ ⃗F=m ⃗a
Can account when m is not constant (rocket propulsion).
∙ Conservation of momentum: For many particle systems, consider N, masses mi and momenta ⃗Pi (i=1,…,N). Let ⃗Fiext be the external force on the ith particle, and ∫¿
⃗Fi¿ be the interaction force of jth on the ith:
⃗F j i∫¿=d ⃗P
dt→ ⃗Ptot=∑i=1 N
⃗Ptot 0=⃗Ptot f
∙ Onedimensional elastic collision (key equations): No external forces => ⃗Ptot 0=⃗Ptot f
m1v1i−m2v2i=−m1v1 f +m2v2 f
→ m1( v1i+v1 f)=m2(v2 f +v2i)
Elastic => Ktot 0=Ktot f
2m1v1i2+12m2v2i2 =12m1v1 f
m1 ( v1i2 −v1 f
2+12m2v2 f 2
∙ Onedimensional inelastic collision:
Momentum is still conserved
If objects stick together after the collision it´s a totally inelastic collision Since KE is not conserved, write ΔK=K f−Ki=Q<0 since ( K f<Ki ) => K f=Ki+Q
Conservation of energy
Given a force ⃗F to a particle of mass m, we have: r f⃗F d ⃗L
∙ Work done by ⃗F :W =∫
∙ Kinetic energy of a particle K=12m v2
∙ If ⃗F is conservative, the potential energy U is F=−dU dx
Elastic: Ue=12R Δ x2 (where R is a constant)
∙ Being E=W +U the total mechanical energy and the ΔE=0 , then KF +UF=K0+U 0
∙ With nonconservative forces KF +UF=K0+U 0+Wnc Momentum ⃗P
Conservation of momentum: ⃗Ptot 0=⃗Ptot f
Onedimensional elastic collision
Momentum: m1 ( v1i+v1 f)=m2(v2 f +v2i)
Kinetic energy: m1 ( v1i2 −v1 f
Onedimensional inelastic collision
Momentum is conserved
Kinetic energy is not conserved: K f=Ki+Q where Q<0