Problems seeing the calculator? Click here to refresh the page. 1. 17N of force is applied over 14.7 metres using a lever. If the load arm is 3 metres long, what is the mass of the box rounded to the nearest tenth of a kilogram? Answer: IMA = Din/Dout IMA = 14.7 m/3 m IMA = 4.9
MA = IMA
Fout = Fin x MA Fout = 17N x 4.9 Fout = 83.3M
Mass = Force/9.8 Mass = 8.5 kg
2. 416.5N of force is used to lift a 17 kg box using a class 1 lever (fulcrum in the middle). If the effort arm is 3.2 metres long, how long is the lever rounded to the nearest tenth of a metre? Answer: Force = Mass x 9.8 Force = 17 kg x 9.8 Force = 166.6N
MA = Fout/Fin MA = 166.6N/416.5N MA = 0.4
IMA = MA
Dout = Din/IMA Dout = 3.2 m/0.4 Dout = 8 m
Length = Din + Dout Length = 3.2 m + 8 m Length = 11.2 m
3. A 26 kg box is raised up a ramp 0.5 metre long and 5 metres wide with a force of 98N. What is the volume of the ramp rounded to the nearest tenth of a cubic metre (assuming the ramp is a triangular prism)? Answer: Force = Mass x 9.8 Force = 26 kg x 9.8 Force = 254.8N
MA = Fout/Fin MA = 254.8N/98N MA = 2.6
IMA = MA
Din = Dout x IMA Din = 0.5 m x 2.6 Din = 1.3 m
Base = square root (c² - a²) Base = square root (1.3² - 0.5²) Base = square root (1.69 - 0.25) Base = square root (1.44) Base = 1.2 m
Volume = (Base x Height x Width)/2 Volume = (1.2 m x 0.5 m x 5 m)/2 Volume = 1.5 m³
4. A machine raises a box 15 metres high with a force of 400N over 2.5 metres with an efficiency of 88.2%. What is the mass of the object rounded to the nearest tenth of a kilogram? Answer: Wi = F x d Wi = 400N x 2.5 m Wi = 1000J
Wo = Wi x (ME/100) Wo = 1000J x (88.2/100%) Wo = 1000J x 0.882 Wo = 882J
F = Wo/d F = 882J/15 m F = 58.8N
Mass = Force/9.8 Mass = 58.8N/9.8 Mass = 6 kg
5. A ramp 10.1 metres long with a base of 9.9 metres is used to lift a load of 81 kg. If the mechanical advantage is 2.05 less than the ideal mechanical advantage, how much energy is wasted rounded to the nearest tenth of a percent? Answer: Dout =square root (c² - b²) Dout =square root (10.1² - 9.9²) Dout =square root (102.01 - 98.01) Dout =square root (4) Dout =2 m
IMA = Din/Dout IMA = 10.1 m/2 m IMA = 5.05
MA = IMA - 2.05 MA = 5.05 - 2.05 MA = 3
Force = Mass x 9.8 Force = 81 kg x 9.8 Force = 793.8N
Fin = Fout/MA Fin = 793.8N/3 Fin = 264.6N
Wo = F x d Wo = 793.8N x 2 m Wo = 1587.6J
Wi = F x d Wi = 264.6N x 10.1 m Wi = 2672.46J
ME = (Wo/Wi) x 100% ME = (1587.6J/2672.46J) x 100% ME = 0.594 x 100% ME = 59.4%
Energy Wasted = 100% - ME Energy Wasted = 100% - 59.4% Energy Wasted = 40.6% Too hard? Try the Practice Questions.
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Problems seeing the calculator? Click here to refresh the page.
1. 17N of force is applied over 14.7 metres using a lever. If the load arm is 3 metres long, what is the mass of the box rounded to the nearest tenth of a kilogram?
Answer:
IMA = Din/Dout
IMA = 14.7 m/3 m
IMA = 4.9
MA = IMA
Fout = Fin x MA
Fout = 17N x 4.9
Fout = 83.3M
Mass = Force/9.8
Mass = 8.5 kg
2. 416.5N of force is used to lift a 17 kg box using a class 1 lever (fulcrum in the middle). If the effort arm is 3.2 metres long, how long is the lever rounded to the nearest tenth of a metre?
Answer:
Force = Mass x 9.8
Force = 17 kg x 9.8
Force = 166.6N
MA = Fout/Fin
MA = 166.6N/416.5N
MA = 0.4
IMA = MA
Dout = Din/IMA
Dout = 3.2 m/0.4
Dout = 8 m
Length = Din + Dout
Length = 3.2 m + 8 m
Length = 11.2 m
3. A 26 kg box is raised up a ramp 0.5 metre long and 5 metres wide with a force of 98N. What is the volume of the ramp rounded to the nearest tenth of a cubic metre (assuming the ramp is a triangular prism)?
Answer:
Force = Mass x 9.8
Force = 26 kg x 9.8
Force = 254.8N
MA = Fout/Fin
MA = 254.8N/98N
MA = 2.6
IMA = MA
Din = Dout x IMA
Din = 0.5 m x 2.6
Din = 1.3 m
Base = square root (c² - a²)
Base = square root (1.3² - 0.5²)
Base = square root (1.69 - 0.25)
Base = square root (1.44)
Base = 1.2 m
Volume = (Base x Height x Width)/2
Volume = (1.2 m x 0.5 m x 5 m)/2
Volume = 1.5 m³
4. A machine raises a box 15 metres high with a force of 400N over 2.5 metres with an efficiency of 88.2%. What is the mass of the object rounded to the nearest tenth of a kilogram?
Answer:
Wi = F x d
Wi = 400N x 2.5 m
Wi = 1000J
Wo = Wi x (ME/100)
Wo = 1000J x (88.2/100%)
Wo = 1000J x 0.882
Wo = 882J
F = Wo/d
F = 882J/15 m
F = 58.8N
Mass = Force/9.8
Mass = 58.8N/9.8
Mass = 6 kg
5. A ramp 10.1 metres long with a base of 9.9 metres is used to lift a load of 81 kg. If the mechanical advantage is 2.05 less than the ideal mechanical advantage, how much energy is wasted rounded to the nearest tenth of a percent?
Answer:
Dout = square root (c² - b²)
Dout = square root (10.1² - 9.9²)
Dout = square root (102.01 - 98.01)
Dout = square root (4)
Dout = 2 m
IMA = Din/Dout
IMA = 10.1 m/2 m
IMA = 5.05
MA = IMA - 2.05
MA = 5.05 - 2.05
MA = 3
Force = Mass x 9.8
Force = 81 kg x 9.8
Force = 793.8N
Fin = Fout/MA
Fin = 793.8N/3
Fin = 264.6N
Wo = F x d
Wo = 793.8N x 2 m
Wo = 1587.6J
Wi = F x d
Wi = 264.6N x 10.1 m
Wi = 2672.46J
ME = (Wo/Wi) x 100%
ME = (1587.6J/2672.46J) x 100%
ME = 0.594 x 100%
ME = 59.4%
Energy Wasted = 100% - ME
Energy Wasted = 100% - 59.4%
Energy Wasted = 40.6%
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