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ENGINEERING, PPT 2024 HSC 24c

An escalator is used to move people from the first floor to the second floor. The second floor is located 5.2 metres above the first floor. The average person's mass is 64.8 kg .

Determine the power requirement of the escalator in order to move 10 people from the first floor to the second floor in one minute.   (3 marks)

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\(561.6\ \text{W}\)

Show Worked Solution

\(W(\text{1 passenger)}\ = mgh = 64.8 \text{kg}\ \times 10 \text{m/s}^{2}\ \times 5.2\ \text{m}\ = 3369.6\ \text{J} \)

\(W(\text{10 passengers)}\ = 10 \times 3369.6 = 33\ 696\ \text{J}\)

\(P=\dfrac{W}{t} = \dfrac{33\ 696\ \text{J}}{60\ \text{s}} = 561.6\ \text{W}\)

ENGINEERING, PPT 2016 HSC 23c

A heavy vehicle is travelling on a level section of the highway at a constant velocity of 90 km/h while experiencing a resistance to motion of 18 kN.

Calculate the power required to maintain this velocity.   (3 marks)

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`450\ text{kW}`

Show Worked Solution
`text{Velocity}` `=90\ text{km/h}`  
  `=(90\ 000)/(60 xx 60)\ text{m/sec}`  
  `=25\ text{m/sec}`  

 

`P` `=Fxxs/t`  
  `=F xx v`  
  `=18\ 000 xx 25`  
  `=450\ text{kW}`  

ENGINEERING, PPT 2019 HSC 18 MC

A scooter shock absorber is compressed from 120 mm to 90 mm when an average compressive force of 400 N is applied.
 

What is the energy stored in this shock absorber?

  1. 12 J
  2. 36 J
  3. 48 J
  4. 120 J
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`A`

Show Worked Solution

`text{30 mm = 0.03 m}`

`W` `= Fs`  
  `= 400 xx 0.03`  
  `= 12\ text{J}`  

 
`=>A`


♦ Mean mark 50%.

ENGINEERING, PPT 2020 HSC 23b

A diagram of the front steering system of a bicycle is shown.
 


 

  1. A rider leans forward so that 55% of their mass is exerted as a force, as shown in the diagram. A turning moment of 132.19 Nm is generated about the front axle.
  2. What is the mass of the rider?   (3 marks)
  1. In overcoming the resistance due to gravity, drag and rolling resistance, a rider applies a force of 500 N on the pedals. This allows the rider to cover, in 5 seconds, a distance of 20 metres from the starting point.
  2. How much power was produced by the rider?  (2 marks)
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i.   `95\ text{kg}`

ii.  `2000\ text{W}`

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i.  
         

`sin20°` `= d/740`  
`d` `= 740 xx sin20°`  
  `= 253.09\ text{mm}`  

 

`M` `= Fxxd`  
`F` `= 132.19/0.25309`  
  `= 522.5\ text{N}`  

 

`55\text{%} xx F_(text{total})` `= 522.5\ text{N}`  
`F_(text{total})` `= 522.5/0.55`  
  `= 950\ text{N}`  

 

`:. M_(text{rider})` `= 950/10`  
  `= 95\ text{kg}`  

♦ Mean mark 47%.

ii.   `F=500\ text{N}, \ s=20\ text{m},\ t=5\ text{sec}`

`P` `= Fxxv`  
  `= F xx s/t`  
  `= 500 xx 20/5`  
  `= 2000\ text{W}`  

ENGINEERING, PPT 2021 HSC 22d

A skateboard rider is at rest at the top of a slope, indicated by point `A` in the diagram.
 

The combined mass of the rider and skateboard is 80 kg.

Calculate the speed of the rider, in metres per second, at point `B` if the average work done by friction against the skateboard is 55 N.    (4 marks)

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`text{Speed} = 14.57\ text{m/s}`

Show Worked Solution

`m= 80\ text{kg}, \ f= 55\ text{N}`

`PE` `= mgh`  
  `= 80xx10xx27.8`  
  `= 22\ 240\ text{J}`  

  
`text{Losses to friction (W):}`

`W` `= fs`  
  `= 55xx250`  
  `= 13\ 750\ text{J}`  

 

`text{Energy used}` `= PE−text{Frictional losses}`  
  `= 22\ 240−13\ 750`  
  `= 8490\ text{J}`  

 
`text{Calculate speed at point B:}`

`1/2×m×v^2` `= 8490`  
`0.5×80×v^2` `= 8490`  
`v^2` `= 212.25`  
`:.v` `= 14.57\ text{m/s}`  

♦ Mean mark 43%.