Final 17 - Springs and Tables

Given:

• $\mbox{A 1 kg mass is connected to a spring with a constant of 350 newtons/meter.} $
• $\mbox{The spring is compressed 25 centimeters and let go from 2 meters.}$

Question:

• $\mbox{How far away will the mass impact the floor?} $

Rationale:

• $PE_{spring} = KE_{mass} $
• $.5 K x^2 = .5 m V_x^2   $ 
• $V_x^2 = \dfrac{(.5 * K * x^2 )}{(.5 * m)}$
• $t_y^2 = \dfrac{2 * \Delta{y}}{g}   $
• $t_y  = t_x $
• $\Delta{x} = V_x * t_x $
• $\Delta{x} = \sqrt{\dfrac{(.5 * K * x^2 )}{(.5 * m)}} *  \sqrt{\dfrac{2 * \Delta{y}}{g}}    $

Calculate:

• $\Delta{x} = \sqrt{\dfrac{(.5 * 350 * .25^2 )}{(.5 * 1)}} *  \sqrt{\dfrac{2 * 2}{10}}    $ 
• $\Delta_{x} = \sqrt{10.9375 * .5} *  \sqrt{.4} = 4.6770717 * .632455 = 2.9580374$ 
• $\Delta_{x} \approx 2.96 \; \mbox{meters}$

References:

• $ $

Final 16 - Bikes and Buses

Final 15 - Gravity and Electricy

Given:

• $\mbox{A mass of 8 kg and a charge of 0.1 coulombs}$
• $\mbox{Is dropped from a height of 1.2 meters}$
• $\mbox{In the continuing presence of an  electric field of 100 newtons/coulomb.} $

Question:

• $\mbox{What distance d does the mass hit the floor?} $

Rationale:

• $\Delta{y} = 0.5 g t^2 $
• $ t^2 = \dfrac{2 \Delta{y} }{g} $
• $F_{electric} = F_x = m a_x $
• $E q =  m a_x $
• $a_x = \dfrac{(E q)}{m} $
• $t_y^2 = t_x^2 $
• $\Delta{x} = .5 (\dfrac{(E q)}{m})  \dfrac{2 \Delta{y} }{g} $

Calculate:

• $\Delta{x} = .5 (\dfrac{(100 * .1)}{8})  \dfrac{2 * 1.2 }{10} = 0.15 \; \mbox{meters}$ 

References:

• $ $

Final 14 - Pushing A Car

Given:

• $\mbox{A car is being pushed by three equal force with one on each side at 30 degrees and one in the rear.} $
• $\mbox{The 1000 kg car is being accelerated at 0.5 m/s/s by  the efforts of these three.}   $

Question:

• $\mbox{What force are each exerting on the car?} $

Rationale:

• $F = F_1 = F_2 = F_3 $
• $F_y = F \cos\alpha \; \mbox{(and there are two of these forces)} $
• $F_{total} = F + (2 F \cos\alpha)   $
• $F_{car} = m a = F + (2 F \cos\alpha)   $
• $ F (1+(2 \sin\alpha) = m a $
• $ F = \dfrac{( m a)}{(1+(2 \cos\alpha)} $

Calculate:

• $F = \dfrac{(100 * .5)}{(1 + (2 * .866025)} = \dfrac{500}{2.73205} = 183.012756 \approx 183$

References:

• $ $

Final 13 - Double Inclined Plane

Given:

• $\mbox{Two masses connected by a rope and pully} $
• $\mbox{mass 1 is on a plane inclined at an angle alpha} $
• $\mbox{mass 2 is on a plane inclined at an angle beta} $
• $a = g \frac{M_{unknown}\sin{\beta}  {(operator)}  M_{unknown}\sin{\alpha}}{M_1 ({operator}) M_2}   $

Question:

• $What are subscripts and operators? $

Rationale:

• $M_1 \; \mbox{is associated with angle alpha} $
• $M_2 \; \mbox{is associated with angle beta} $
• $a = g \frac{M_2 \sin{\beta}  {(operator)}  M_1 \sin{\alpha}}{M_1 ({operator}) M_2}   $
• $\mbox{The denominator operator cannot be multiply or divide}$
• $\mbox{Since if either mass is zero, then the equation  would be indeterminate}   $
• $\mbox{So the operator must be + or - }$
• $\mbox{But it cannot be minus,  since it would indeterminate is the masses are equal}$
• $\mbox{Therefore the denominator operator is plus.}$
• $\mbox{If the masses are equal and the angles are equal, }$
• $\mbox{Then the numerator must be zero}$
• $\mbox{Thus the numerator operator must be minus.}$
• 
  

Calculate:

• $ $

References:

• $ $

Final 12 - Water Clock

Given:

• $\mbox{In one time interval, a ball rolls down the incline 0.8 meters.} $

Question:

• $\mbox{What interval does the ball travel in the second and third time intervals?} $

Rationale & Calculate:

• $\Delta{x} \varpropto t^2 $
• $\Delta{x-1} = 1 * K $
• $\Delta{x-2} = 4 * K$ 
• $\Delta{x-3} = 9 * K $ 
• $ K = 0.8 \; \mbox{since at t-1, x-1 is 0.8} $
• $\Delta{x-2} = 3.2 $ 
• $\Delta{x-3} = 7.2 $ 
• $ Interval_{2-1} = 3.2 - 0.8 = 2.4 $
• $ Interval_{3-2} = 7.2 - 3.2 = 4.0 $

References:

• Unit 2A-8 of PH-100

Final 11 - Dead Reckoning

Given:

• $\mbox{A ship travels:}$
• $\mbox{East at 12 km/hour for 2 hours} $
• $\mbox{Then south at 20 km/hour for 1 hour}$
• $\mbox{Then east at 15 km/hour for 3 hours}$
• $\mbox{Then northeast into port at 8 km/hour for 1 hour.} $

Question:

• $\mbox{What is the straight line distance traveled to port ? }$

Rationale:

• $\mbox{Segment-distance}  = velocity * time $
• $East_{distance} = \Delta{x} $
• $South_{distance} = \Delta{y} $
• $-\Delta{y_{northeast} = \Delta{x_{northeast}} = \sin{45^{\circ}}} * Northeast_{distance} $
• $\mbox{Straight-line-distance} = \sqrt{(\Delta_{x-total})^2 + (\Delta_{y-total})^2 } $

Calculate:

• $East_{distance} = ((12) (2.5)) + ((15) (3)) =  75 $
• $South_{distance} = (20) (1) = 20 $
• $ -\Delta{y_{northeast}}  = \Delta{x_{northeast}} = (0.707106781) (8) (1) = 5.656854 $
• $\Delta_{x} = 75 +  5.656854 = 80.656854 $
• $\Delta_{y} = 20 - 5.656854 = 14.343146 $
• $\mbox{Straight-line-distance} = \sqrt{(80.656854)^2 + (14.343146)^2} $
• $\mbox{Straight-line-distance} = \sqrt{(6505.5280 + 205.7258 )} = \sqrt{6711.2538}  $
• $\mbox{Straight-line-distance} =81.9222 \approx 81.9 $

References:

• $ $

Final 10 - Falling Water

Given:

• $\mbox{Water is flowing down a wall of 1.5 meters and reaches a velocity of 4 meters/second at the bottom.} $

Question:

• $\mbox{What is the work done by friction of a 1 gram water drop to retard its falling freely?} $

Rationale:

• $V^2_{water-at-bottom} = 2 a_{retarded-from-normal-g} {height} $
• $a_{retarded-from-normal-g} = \dfrac{V^2_{water-at-bottom}}{(2 height)}   $
• $a_{friction} = g - a_{retarded-from-normal-g}  $
• $W_{friction} = (m) (a_{friction}) {(height)}   $

Calculate:

• $a_{retarded-from-normal-g} = \dfrac{(4)^2}{(2 * 1.5)} = 5.33333 $
• $a_{friction} = 10 - 5.33333 = 4.66667 $
• $ W_{friction} =  (1 * 10^{-3})(4.66667)(1.5) = 7 * 10^{-3} = .007 $

References:

• $ $

Final 9 - Strange Planet

Given:

• $\mbox{Length of pendulum is 12 meters and it took 45 seconds for it to swing back and forth.} $

Question:

• $\mbox{What is the force of gravity on this planet?} $

Rationale:

• $T = 2 \pi \sqrt{\dfrac{l}{g}} $

Calculate:

• $\dfrac{45}{10} = (2) \dfrac{22}{7} \sqrt{\dfrac{12}{g}} $
• $  (4.5) \dfrac{7}{44} = \sqrt{\dfrac{12}{g}} $
• $ 0.71590 = \sqrt{\dfrac{12}{g}} $
• $ 0.51251281 = \dfrac{12}{g}  $
• $ g = \dfrac{12}{ 0.51251281} = 23.4140489 \approx 23.4 $

References:

• $ $

Final 8 - Cable Car Power

Given:

• $ \mbox{A 2,000 kg cable car on an alpha 10 degree slope.} $

Question:

• $\mbox{How much power does it take to move the trolley at 25 km/hour?} $

Rationale:

• $25 \dfrac{km}{hour} = 25 \dfrac{1000 \mbox{meters-per-km}}{3600 \mbox{seconds-per-hour}} $
•  $25 \dfrac{km}{hour} =6.944444 \; \mbox{meters per second} $
• $ P = F_{parallel} V = m g \sin\alpha V$

Calculate:

• $P = (2000) (10) (.173648178) (6.9444) = 24,117.648146 $
•  $ P = \approx 24,117 $

References:

• $ $

Final 7 - Positive Latitude

Given:

• $\mbox{A 1 m rod casts a 1 m shadow at noon on the summer solstice.} $
• $ \mbox{At the equator, the sun's rays strike the earth vertically at 23.5 degrees north.}$

Question:

• $\mbox{What is the latitude at the measurement point?}$

Rationale:

• $\mbox{The triangle with 1 m on both sides is a 45 degree triangle.}$
• $\mbox{90 - The angle at the measurement point} + \mbox{The angle at the equator} = latitude$

Calculate:

• $ latitude = 45 + 23.5 = 68.5 $

References:

• Calculating Latitude

Final 6 - Balloon Balance

Given:

• $\mbox{In an electrical field of 10,000 N/C, a mass of 30 grams is charged.} $

Question:

• $\mbox{How large must that charge be to suspend the mass neither moving up nor down?} $

Rationale:

• $F_{gravity} = F{electrical} $
• $m a = E q   $
• $q = \dfrac{m a}{E}   $

Calculate:

• $q = \dfrac{(30 * 10^{-3}) (10)}{10^4} = 3 * 10^{-5} $

References:

• $ $

Final 5 - Trolley

Given:

• $\mbox{A 1,000 kg trollley on a slope of 10 degrees.} $

Question:

• $\mbox{What force is needed to cause the trolley remain stationary ?} $

Rationale:

• $ F_{parallel} = m g \sin\alpha $

Calculate:

• $ F_{parallel} = (1000) (10) \sin{10_{circ}} = (1000) (10) (.173648) $
• $ F_{parallel} = 1,736.48 \; \approx \; 1,736.5 $

References:

• $ $

Final 4 - Climbing Stairs

Given:

• $\mbox{A 65 kg person climbs some 6 meter stairs.}$

Question:

• $\mbox{How much work is done?} $

Rationale:

• $ W = m g h $

Calculate:

• $ W =  (65) (10) (6) = 3,900 \; \mbox{joules} $

References:

• $ $

Final 3 - Reducing the Period

Given:

• $\mbox{Mass m attached to a spring with spring constant K.} $

Question:

• $\mbox{What size mass should you attach to the spring to double the period of oscillation?} $

Rationale:

• $ T_{period} = 2 \pi \sqrt{\dfrac{m}{k}}$
• $ 2 T_{period} = 2 \pi 2 \sqrt{\dfrac{m}{k}}$
• $ 2 T_{period} = 2 \pi \sqrt{\dfrac{4 m}{k}}$

Calculate:

• $ \mbox{Answer is it takes 4m to double the period.} $

References:

• $ $

Final 2 - Height of the Church

Given:

• $\mbox{Observer #1 is at ground level} $
• $\mbox{Sees the top of  head of Observer #2 and top of church in line.}   $
• $\mbox{Observer #2} = 1 \; \mbox{meter tall}   $
• $\mbox{Meters from Observer #1} = 2   $
• $ \mbox{Meters from church} = 90  $

Question:

• $\mbox{How tall is the church?} $

Rationale:

• $\mbox{We have two similar triangles one within the other.} $
• $ \mbox{A small one: 1 meters vertical by  2 meters horizontal describing an angle} = \theta   $
• $ \mbox{A large one: the height of church vertical by 92 meters horizontal  describing an angle} = \theta $
• $ \dfrac{vertical_{small}}{horizontal_{small}} = \tan\theta   $
• $ \dfrac{vertical_{church}}{horizontal_{church}} = \tan\theta $

Calculate:

• $\tan\theta = \dfrac{1}{2} \; \mbox{using the small triangle} $
• $ \dfrac{vertical_{church}}{horizontal_{church}} = \tan\theta $
• $\dfrac{vertical_{church}}{92} = \tan\theta = 0.5   $
• $ vertical_{church} = 0.5 * 92 = 46 \; \mbox{meters}   $

References:

• Similar Triangles

Rosa's Summary of Class Formulae

for uniform motion (a=0):

• v=v0 and
  
• Δs=v0×Δt
  

for uniformly accelerated motion (a=constant≠0):

• v=v0+aΔt,
  
• Δs=v0×Δt+12a×Δt2 and
  
• v2=v20+2×a×Δs
  

for the gravitational force (weight - always in the vertical down direction):

• P=m×g

for the elastic force (always opposite elastic displacement):

• Felas=−kΔx, k is the elastic constant

for the electric force (attractive for opposite sign charges and repulsive for equal sign charges):

• Fe=keq1q2r2, ke is the electric constant

for the second law of Newton, in the x direction (for example):

• Funbalanced,x=m×ax

for work done by a constant force (careful, Fparallel may be negative and its the force component in the displacement direction):

• W=Fparallel×displacement

for average power:

• P=WΔt

for instantaneous power

• P=F⃗ ⋅v⃗ 

for the conservation of energy:

• Etotalinitial=Etotalfinal

for the kinetic, potential and total energy:

• K=12mv2
  
• Ugravitacional,approximate=mgh
  
• Ugravitacional,exact=−Gm.Mr
  
• Uelast=12kΔx2
  
• Uelectric=keq1q2r1,2+keq1q3r1,3+keq2q3r2,3+...
  
• Etotal=K+Ugravitacional+Uelast+Uelectric
  

for simple harmonic oscilation:

• a=−ω2x
  
• ω=2πT=ΔθΔt
  
• pendulum: ω2=km
  
• spring-mass: ω2=gl
  

for the gravitic field created by a mass M, acting on a mass m:

• g=GMr2
  
• weight=FG=m×g
  

for the elecric field E acting on a charge q:
Fe=q×E