Force_Work_ Mechanical Advantage_Efficiency Practice Problems

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Force/Work/ Mechanical Advantage/Efficiency
Practice Problems
   1. What is the formula for Force? Include the required units under your formula.
      Write the formula in the “triangle” format as well.




   2. What is the formula for mass? For acceleration (gravity)?

       m = F/g
                                                         a = F/m

   3. What is the formula for work? Include the required units under your formula.
      Write the formula in a “triangle” format as well.




   4. What is a second formula for Force? For distance?

                                                         d = W/F
       F = W/d
   5. a. If a student lifts a textbook that is 1200g, how much force does she use?
       Remember to use the proper units!
Given: m=1200g         **must convert this             Calculations: F=ma
        = 1.2kg        to proper units!                               = 1.2kg x 10m//s2
                  2
Known: g=10m/s                                                        = 12N
Formula: F=mg                   The student uses 12 N of force to lift the textbook.
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   b. For the above book, how much work does the student do if she moves the book
       5meters.
Given: m=1.2kg (above)             Calculations: W=F.d
       F=12N (calculated above)                   =12N x5m
       d=5 m                                      = 60J
Formula: W = F.d            The student did 60J of work to lift the textbook 5 meters.

   6. Find the force and work done in the following problems:
             a.      mass = 15O kg                F=ma
                     Force = ? 1500N               = 150kg x 10m/s2
                     distance = 40 cm             W=Fd
                     (must convert!) =0.4m         = 1500N x 0.4m
                     Work = ? 600J                  = 600 J

              b.   Mass = 270g                     F=ma
                   convert to kg = 0.27kg           = 0.27kg x 10 m/s2
                   Force = ?                       = 2.7N
                   Distance = 5m           W=Fd
                   Work = ?                  = 2.7N x 5m
                                             = 13.5J
   7. a. How much force would it take to lift a 60kg box onto a truck 2m high? (this is
      LOAD force – the force needed to lift the box WITHOUT a machine)
   Given: m=60kg                           Calculations: F=mg
        d= 2m                                             =60kg x 10m/s2
                                                          =600N
   Formula: F=mg           It would take 600N of force to lift the 60kg box.

   b. How much work would be done? (this is Work OUTPUT – the work needed if you
   do not have a machine)
   Given: FL = 600N             Wo = FL x dL
          d= 2m                   = 600N x 2m
                                 =1200J

   Formula: Wo = FL x dL                    Therefore, 1200J of work would be done.

       c. If it takes 200N to push the same box up a 10 m ramp and onto the truck, how
           much work would it take? (this would be the work input        ) – the work
           required if you use a machine.
       Given:
       F = 200N
       d=10m                         It would take 2000J to use this machine.
       Formula: WI = 200N x 10 m
                     = 2000J
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   d. What is the mechanical advantage of the ramp?
   Formula: M.A. = FL/FO
                 = 600N/200N
                 =3X
    The mechanical advantage of the machine is 3X.

   e. What is the efficiency of the machine?
   Efficiency = WO / WI x 100%
                 = 1200J/2000J x100%                    The efficiency is 60%
                 = 60%

8. If the effort force is 15N and the load force is 30N, what is the mechanical
   advantage of the machine?
Given: FE = 15N
            FL = 30N                      The Mechanical Advantage of the
                                                   Machine is 2X.
Formula: M.A. = FL / FE
              = 30N/15N
            = 2X
   9.      effort force = 10N            FL      = mg
           m=3kg                                 = 3kg x 10m/s2
           Load force = ?                        = 30N
           Mechanical advantage =?

           M.A. = FL / FE
                = 30N/10N
           M.A. = 3X

   10.     Work Input = 90J
           Work Output = 30J
           Efficiency =?
           Efficiency    = WO/WI x 100%
                         =30J/90J x 100%
                         = 33%
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 11. What is the efficiency of a machine that lifts a 1kg weight 5 m up to a ledge. To use
this machine, you exert 5 N while moving 15 m?
givens:
m=1kg                                 FL      = ma
d=5m(output)                                  = 1kg x 10m/s2
FE= 5N                                        = 10N
d=15m (input)           WI     =FE x d (input)         WO     = FL x d (output)
                        WI     = 5N x 15m              WO     = 10N x 5m
                        WI     =75N                    Wo     =50J
Formulas:
F=ma                                  Efficiency       = WO / WI x 100%
W=Fd                                                   = 50J/75J x 100%
Efficiency = WO / WI x 100%                            = 66%

                Therefore, the efficiency of the machine is 66%.

     12. A 10 kg mass needs to be lifted onto a truck that is 1.5 m from the ground. A
     ramp, 10m long, was used to get the mass onto the truck.

        a. What was the mechanical advantage for this inclined plane, assuming the
           effort force was only 37.5N?

Given:
m=10kg
d=1.5m (output)                       FL      = mg
d=4m (input)                                  = 10kg x 10m/s2
FE = 37.5N                                    = 100N

Formulas:                             M.A. = FL / FE
F=ma                                       = 100N/37.5N
M.A. = FL / FE                             =2.7X

        The mechanical advantage for this machine is 2.7X.

                a. What is the efficiency of the above machine?
WI      = FE x d (input)                             WO      = FL x d(output)
        = 37.5N x 4m                                         =100N x 1.5m
        = 150J                                               =150J
                      Efficiency    = WO / WI x 100%
                                    =150J / 150J x 100%
                                    = 100%
Therefore, the efficiency of the machine is 100%!!! Impossible, since no machine can
be 100% efficient. I made a mistake when I was making up the question!

				
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