Problem 1 Output Work, Input Work, Slope

Complete the following: In your journal,  Draw a picture of an inclined plane  Place the following labels and values in the correct locations: the correct locations:  Load Force = 20 Newton’s  Load Distance = 3 m  Effort Force = 15 Newton’s  Effort Distance = 5 m  Rise = 3 m  Run = 4m Calculate Output Work? Calculate Input Work? What is the slope of the incline?

< Run---4 m  Slope = Rise Run Slope = 3 (rise) 4 (run) Slope =.75 Slope = Rise Run < Rise m  Slope

< Run---4 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---3 m  Output Work = Load Force x Load Distance OW = 20 x 3 = 60 (LF) (LD) Joules OW = 60 Joules OW = LF x LD OW = 60 Joules Output Work  Load Force Newton’s---- 

< Run---4 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---3 m  Input Work = Effort Force x Effort Distance IW = 15 x 5 = 75 (EF) (ED) Joules IW = 75 Joules OW = LF x LD OW = 60 Joules Input Work  Load Force Newton’s---  <------Effort Distance---5 m-----  <------Effort Force—15 Newton’s--  IW = EF x ED IW = 75 Joules

Problem 2 IW, OW, Slope, IMA, AMA, Efficiency

Complete the following: In your journal,  Draw a picture of an inclined plane  Place the following labels and values in the correct locations: the correct locations:  Load Force = 15 Newton’s  Load Distance = 6 m  Effort Force = 11 Newton’s  Effort Distance = 10 m  Run = 8m Calculate Output Work? Calculate Input Work? What is the slope of the incline? What is the IMA? What is the AMA? What is the efficiency?

< Run---8 m  Slope = Rise Run Slope = 6 (rise) 8 (run) Slope =.75 Slope = Rise Run <---Rise— (same as Load Distance) m  Slope

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  Output Work = Load Force x Load Distance OW = 15 x 6 = 90 (LF) (LD) Joules OW = 90 Joules OW = LF x LD OW = 90 Joules Output Work  Load Force Newton’s---- 

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  Input Work = Effort Force x Effort Distance IW = 11 x 10 = 110 (EF) (ED) Joules IW = 110 Joules OW = LF x LD OW = 90 Joules Input Work  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—11 Newton’s--  IW = EF x ED IW = 110 Joules

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  IMA = Effort Distance ÷ Load Distance IMA = 10 ÷ 6 = 1.67 (ED) (LD) IMA = 1.67 OW = LF x LD OW = 90 Joules IMA  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—11 Newton’s--  IW = EF x ED IMA = 1.67 IW = 110 Joules IMA = ED ÷ LD

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  AMA = Load Force ÷ Effort Force AMA = 15 ÷ 11 = 1.36 (LF) (EF) AMA = 1.36 OW = LF x LD OW = 90 Joules AMA  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—11 Newton’s--  IW = EF x ED IMA = 1.67 IW = 110 Joules IMA = ED ÷ LD AMA = LF ÷ EF AMA = 1.36

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  Efficiency = (OW ÷ IW) x 100 Efficiency = (90 ÷ 110) =.82 (OW) (IW).82 x 100 = 82 % Efficiency = 82% OW = LF x LD OW = 90 Joules Efficiency  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—11 Newton’s--  IW = EF x ED IMA = 1.66 IW = 110 Joules IMA = ED ÷ LD AMA = LF ÷ EF AMA = 1.36 Eff = OW ÷ IW x 100 Eff = 82%

Problem 3 IW, OW, Slope, IMA, AMA, Efficiency

Complete the following: In your journal,  Draw a picture of an inclined plane  Place the following labels and values in the correct locations: the correct locations:  Load Force = 120 Newton’s  Load Distance = 12 m  Effort Force = 80 Newton’s  Effort Distance = 22.5 m  Run = 20 m Calculate Output Work? Calculate Input Work? What is the slope of the incline? What is the IMA? What is the AMA? What is the efficiency?

< Run---20 m  Slope = Rise Run Slope = 12 (rise) 20 (run) Slope =.6 Slope = Rise Run <---Rise— (same as Load Distance) m  Slope

< Run---20 m  Slope =.6 Slope = Rise Run < Rise m  Load Distance---12 m  Output Work = Load Force x Load Distance OW = 120 x 12 = 1440 (LF) (LD) Joules OW = 1440 Joules OW = LF x LD OW = 1440 Joules Output Work  -----Load Force Newton’s---- 

< Run---20 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---12 m  Input Work = Effort Force x Effort Distance IW = 80 x 22.5 = 1800 (EF) (ED) Joules IW = 1800 Joules OW = LF x LD OW = 1440 Joules Input Work  Load Force Newton’s-  <------Effort Distance m-----  <------Effort Force—80 Newton’s--  IW = EF x ED IW = 1800 Joules

< Run---20 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---12 m  IMA = Effort Distance ÷ Load Distance IMA = 22.5 ÷ 12 = 1.88 (ED) (LD) IMA = 1.88 OW = LF x LD OW = 1440 Joules IMA  -----Load Force Newton’s---  <------Effort Distance m-----  <------Effort Force—80 Newton’s--  IW = EF x ED IMA = 1.88 IW = 1800 Joules IMA = ED ÷ LD

< Run---20 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---12 m  AMA = Load Force ÷ Effort Force AMA = 120 ÷ 80 = 1.5 (LF) (EF) AMA = 1.5 OW = LF x LD OW = 1440 Joules AMA  Load Force Newton’s---  <------Effort Distance m-----  <------Effort Force—80 Newton’s--  IW = EF x ED IMA = 1.88 IW = 1800 Joules IMA = ED ÷ LD AMA = LF ÷ EF AMA = 1.5

< Run---20 m  Slope =.75 Slope = Rise Run < Rise m  -----Load Distance---12 m  Efficiency = (OW ÷ IW) x 100 Efficiency = (1440 ÷ 1800) =.80 (OW) (IW).80 x 100 = 80 % Efficiency = 80% OW = LF x LD OW = 1440 Joules Efficiency  -----Load Force Newton’s---  <------Effort Distance m-----  <------Effort Force—80 Newton’s--  IW = EF x ED IMA = 1.88 IW = 1800 Joules IMA = ED ÷ LD AMA = LF ÷ EF AMA = 1.5 Eff = OW ÷ IW x 100 Eff = 80%

Problem 4 IW, OW, Slope, IMA, AMA, Efficiency

Complete the following: In your journal,  Draw a picture of an inclined plane  Place the following labels and values in the correct locations: the correct locations:  Load Force = 18 Newton’s  Load Distance = 6 m  Effort Force = 12 Newton’s  Effort Distance = 10 m  Run = 8m Calculate Output Work? Calculate Input Work? What is the slope of the incline? What is the IMA? What is the AMA? What is the efficiency?

< Run---8 m  Slope = Rise Run Slope = 6 (rise) 8 (run) Slope =.75 Slope = Rise Run <---Rise— (same as Load Distance) m  Slope Click to continue

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  Output Work = Load Force x Load Distance OW = 18 x 6 = 108 (LF) (LD) Joules OW = 108 Joules OW = LF x LD OW = 108 Joules Output Work  Load Force Newton’s----  Click to continue

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  Input Work = Effort Force x Effort Distance IW = 12 x 10 = 120 (EF) (ED) Joules IW = 120 Joules OW = LF x LD OW = 108 Joules Input Work  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—12 Newton’s--  IW = EF x ED IW = 120 Joules Click to continue

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  IMA = Effort Distance ÷ Load Distance IMA = 10 ÷ 6 = 1.67 (ED) (LD) IMA = 1.67 OW = LF x LD OW = 108 Joules IMA  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—12 Newton’s--  IW = EF x ED IMA = 1.67 IW = 120 Joules IMA = ED ÷ LD Click to continue

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  AMA = Load Force ÷ Effort Force AMA = 18 ÷ 12 = 1.5 (LF) (EF) AMA = 1.5 OW = LF x LD OW = 108 Joules AMA  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—12 Newton’s--  IW = EF x ED IMA = 1.67 IW = 120 Joules IMA = ED ÷ LD AMA = LF ÷ EF AMA = 1.5 Click to continue

< Run---8 m  Slope =.75 Slope = Rise Run < Rise m  Load Distance---6 m  Efficiency = (OW ÷ IW) x 100 Efficiency = (108 ÷ 120) =.90 (OW) (IW).90 x 100 = 90% Efficiency = 90% OW = LF x LD OW = 108 Joules Efficiency  Load Force Newton’s---  <------Effort Distance---10 m-----  <------Effort Force—12 Newton’s--  IW = EF x ED IMA = 1.66 IW = 120 Joules IMA = ED ÷ LD AMA = LF ÷ EF AMA = 1.5 Eff = OW ÷ IW x 100 Eff = 90% Click to continue