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Solving Physics Problems
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Solving Physics Problems


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A general approach to problem solving in physics. …

A general approach to problem solving in physics.
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  • 1. How to Solve Physics Problems
  • 2. The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www. wsautter .com
  • 3. Books available at: www. wsautter .com Walt’s Books for Free!
  • 4. Solving problems in physics generally requires a few basic but essential steps. Read the question carefully and decide nature of the answer. What are we asked to find? Mass, velocity, displacement, force, etc. Write down the symbol for the answer with a question mark. For example: m = ?, v = ?, d =? or f = ? Reread the question to find out what information is given. Record each bit of information as the problem is read. For example: v o = 2.0 m/s, t = 3.0 sec, v f = 5.4 m/s, etc. Notice that the units are included with each value. Units can often be used to decide the nature of each value even if you are not told directly in the problem what the number value represents. For example: m/s must velocity or speed, newtons must a force, seconds must be time, etc.
  • 5. Unit Systems Unit systems are specified as MKS* (larger metric units), CGS (smaller metric units) and English units. Working with units generally requires us to stay in the Same unit group for all values used in solving a problem. For example: we would not use newtons (an MKS unit) with grams (CGS). We would convert grams to kilograms in order to use it with newtons. Similarly, cm/ sec (CGS) would not be used with meters (MKS), hours would not be used with seconds. The MKS unit system is also called SI units. (System Internationale)
  • 6. Units of Commonly used Systems MKS CGS English Displacement Distance Meters (m) centimeters (cm) feet (ft) Mass Kilogram (kg) gram (g) slug(sg) Seconds (s) Seconds (s) Seconds (s) Time Meter/ sec centimeter/ sec feet/ sec (m/s) (cm/s) (ft/s) velocity speed acceleration Meter/ sec 2 centimeter/ sec 2 feet/ sec 2 (m/s 2 ) (cm/s 2 ) (ft/s 2 ) force newtons (N) dynes(dn) pound (lb)
  • 7. Units of Commonly used Systems (cont’d) MKS CGS English Work energy Kilojoules (Kj) ergs(er) foot pound (ft-lb) power Kilowatt (Kw) watt(w) horsepower (hp) Kilojoules joules calories (Kj) (j) (cal) Heat energy Impulse Newton x sec dyne x sec pound x sec (N x s) (dn x s) (lb x s) momentum Kilogram x m/sec gram x cm/sec slug x ft/sec Kg x m/s g x cm/s sg x ft/s Newton x meter dyne x centimeter foot x pound N x m dn x cm ft x lb torque
  • 8. More Commonly Used Units angles frequency period Angular displacement Angular velocity Angular acceleration Degrees radians revolutions Revolutions per second (rps) hertzs (hz) Seconds / revolution Radians Radians / second Radians / second 2
  • 9. After identifying all the information given in the problem and Deciding on what is to be found, the next step is to select an Equation containing the unknown value. Next, see if the selected equation contains all the variables that are given in the problem. If so, insert the number values in the appropriate spots in the equation and solve. If the data for one of the required variables for solving the equation is missing search the other available equations for one that contains the missing variable and known data. This equation will allow you to find the missing variable value. Calculate its value and insert it into the equation containing the unknown and solve for the answer.
  • 10. Solving a problem using the described method. A car moving at 20.0 meters per second brakes at 3.0 meters per second 2 in 0.11 minutes. What is its stopping distance? Read the problem. What are we looking for? What are the units for the answer? Distance d = ? m In meters! MKS units! What data is given? Write down and label each value with a symbol and proper units Starting velocity V o = 20.0 m/s acceleration a = -3.0 m/s 2 (it’s slowing) final velocity V f = 0 m/s (stops) Time T = 0.5 min (0.11 x 60) = 6.7 s
  • 11. Available Equations Both equations (3) and (4) contain our unknown (d). Since we know Vi , V o , a and t either equation will Work. Try both! Did you get 67 meters? I hope so! (1) V AVERAGE =  s/  t = (V 2 + V 1 ) / (2) V INST. = V ORIGINAL + at (3) d INST = V 0 t + ½ at 2 (4) d i = ½ (Vi 2 – V o 2 ) /a
  • 12. Let’s try another: A 200. newton object slows from 50.0 m/s to rest in 10.0 seconds. What is the braking force applied to the object? Read the problem. What are we looking for? What are the units for the answer? Force in newtons f = ? N What data is given? Write down and label each value with a symbol and proper units Starting velocity V o = 50.0 m/s final velocity V f = 0 m/s (to rest) Time T =10.0 sec Weight W = 200. N Not mass that would Be kilograms
  • 13. Available Equations Only equation (6) contains our unknown (F). We have to use it! But we need m and a! We can use equation (5) find m since we know both w and g (9.8 m/s 2 ) Next we can use equation (2) find a since we know V final , V 0 and t. Now we’ll insert the values found for m and a Into equation (6) and calculate the answer. (1) V AVERAGE =  s/  t = (V 2 + V 1 ) / (2) V final = V ORIGINAL + at (3) d INST = V 0 t + ½ at 2 (4) d i = ½ (Vi 2 – V o 2 ) /a (5) W = m x g (6) F = m x a
  • 14. Did you get 20.4 kg for the mass? Did you get - 5.0 m/s 2 for the acceleration? Remember it’s negative because it’s Slowing! Did you get -102 N for the force? This is a generalized procedure for solving most physics problems. Continue to use it and physics will become much easier.
  • 15. The End