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7/28/2019 Ch 19 Practice Sheet Answers http://slidepdf.com/reader/full/ch-19-practice-sheet-answers 1/4 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)?  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?  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//s 2  Known: g=10m/s 2 = 12N Formula: F=mg The student uses 12 N of force to lift the textbook. 1 m = F/g a = F/m F = W/d d = W/F 

Ch 19 Practice Sheet Answers

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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)?

 

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?

 

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

 Known: g=10m/s2 = 12N 

Formula: F=mg The student uses 12 N of force to lift the textbook.

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m = F/g a = F/m

F = W/d d = W/F 

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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=maconvert 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 W  o = F  L x d  Ld= 2m = 600N x 2m

  =1200J 

Formula: W o = F  L x d  L 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. = F  L /F O 

= 600N/200N 

= 3 X 

  The mechanical advantage of the machine is 3X.

e. What is the efficiency of the machine? Efficiency = W O / W  I  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: F  E = 15N 

F  L = 30N The Mechanical Advantage of the

 Machine is 2X.Formula: M.A. = F  L / F  E 

  = 30N/15N 

= 2X 

9. effort force = 10N F  L = mg 

m=3kg = 3kg x 10m/s2

Load force = ? = 30N 

Mechanical advantage =?

 M.A. = F  L / F  E 

= 30N/10N 

 M.A. = 3X 

10. Work Input = 90JWork Output = 30J

Efficiency =?

 Efficiency = W O /W  I  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 usethis machine, you exert 5 N while moving 15 m?

givens:

m=1kg F   L = ma

d=5m(output) = 1kg x 10m/s2

F  E = 5N  = 10N d=15m (input) W   I  =F  E x d (input)  W O = F  L x d (output)

W  I  = 5N x 15m W  O = 10N x 5m

W  I  =75N W  o =50J 

Formulas:

F=ma Efficiency = W  O / W  I  x 100%

W=Fd = 50J/75J x 100%

 Efficiency = W O / W  I  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. Aramp, 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) F   L = mg 

d=4m (input) = 10kg x 10m/s2

F  E  = 37.5N = 100N  

Formulas: M.A. = F  L / F  E 

F=ma = 100N/37.5N  

 M.A. = F  L / F  E  =2.7X 

The mechanical advantage for this machine is 2.7X.

a. What is the efficiency of the above machine?

W  I  = F  E x d (input) W  O = F  L x d(output)

= 37.5N x 4m =100N x 1.5m

= 150J =150J   Efficiency = W O / W  I  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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