SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

ENGINEERING, PPT 2024 HSC 10 MC

An inclined plane is 5 metres long and 2 metres high. A force of 50 N is required to push a box up along the length of the slope.
 

Assuming no friction and 100% efficiency, what is the mass of the box to the nearest kilogram?

  1. 5
  2. 13
  3. 20
  4. 125
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Energy moving box}\ = 50 \times 5 = 250\ \text{J}\)

\(\text{Energy used = increase in potential energy of box (100% efficiency)}\)

\(m \times g \times h\) \(=250\)  
\(m\) \(=\dfrac{250}{10 \times 2}=12.5\ \text{kg}\)  

 
\(\Rightarrow B\)

♦ Mean mark 40%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-50-Inclined planes

ENGINEERING, PPT 2023 HSC 18 MC

A diagram of a block being pulled by a rope up an inclined plane is shown.
 

Which free body diagram correctly represents the diagram given?
 


 

Show Answers Only

\(  C \)

Show Worked Solution

\(\Rightarrow  C \)

Filed Under: Mechanics Tagged With: Band 4, smc-3718-50-Inclined planes

ENGINEERING, PPT 2017 HSC 3 MC

A mass `M` sits on a plane inclined at `theta`° to the horizontal.
 

What expression is used to calculate the normal reaction `N` ?

  1. `Mg`
  2. `Mg\ costheta`
  3. `Mg\ sintheta`
  4. `Mg\ tantheta`
Show Answers Only

`B`

Show Worked Solution

`=>B`

Filed Under: Mechanics Tagged With: Band 4, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2018 HSC 20 MC

A box with weight `W` and subject to a friction force `F` is being pulled up an inclined plane by a force `P`.
 

The force `R` is the resultant of which two forces?

  1. `P` and `W`
  2. `P` and `N`
  3. `F` and `W`
  4. `F` and `N`
Show Answers Only

`D`

Show Worked Solution
  • `R` is used as the resultant of `F` and `N`, usually used to calculate `P` and employ the 3 force rule.

`=>D`


♦ Mean mark 42%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-30-Friction, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2017 HSC 24b

A child and sled with a combined mass of 23 kg are being pulled along a horizontal snow-covered surface using a rope.
 

The coefficient of static friction between the sled and the snow is 0.14.

  1. Draw a free-body diagram that indicates the forces acting on the sled. Label the diagram.   (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

  1. Calculate the tension in the rope immediately before the point at which the sled and child begin to move.   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

i.
       

 

ii.   `text{32 N}`

Show Worked Solution
i.
       

Mean mark (i) 57%.

ii.   `text{Weight}\ = mg = 23 xx 10 = 230\ text{N}`     

`text{Solving for}\ T:`

`tan^(-1)(0.14) = 8°`

`text{Using the sine rule:}`

`T/(sin8°)` `=230/(sin92°)`  
`T` `=(230 xx sin8°)/(sin92°)=32.0\ text{N}`  

  
`:.\ text{Tension in the rope = 32 N}`


♦ Mean mark (ii) 40%.

Filed Under: Mechanics Tagged With: Band 4, Band 5, smc-3718-30-Friction, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2020 HSC 14 MC

The diagram shows the forces acting on a block on an inclined plane. The block is held in a stationary position by force `P`.
 

What is the correct equation for the normal force, `N` ?

  1. `N=W cos theta - P sin phi`
  2. `N=W cos theta + P sin phi`
  3. `N=W sin theta - P cos phi`
  4. `N=W sin theta + P cos phi`
Show Answers Only

`A`

Show Worked Solution

  • `P\ sin phi` is the upward force on the block perpendicular to the plane.

  • `Wcos\ theta`  is the downward force on the block perpendicular to the plane.
  • Therefore, `N` is equal to the downward component of the weight force minus the upwards component of force `P`.

`=>A`


♦ Mean mark 50%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2021 HSC 26b

The diagram shows a box which is at rest on an inclined ramp. The length of the ramp is 3 m and one end is raised `h` metres off the ground.
 


 

The coefficient of friction between the ramp and the box is 0.5.

To what height `(h)` must the ramp be raised for the box to just begin to slide?   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`1.34\ text{m}`

Show Worked Solution
`mu` `= 0.5 = tan\ theta`  
`theta`  `= tan^(-1)(0.5)=  26.57°`  

 

`sin(26.57°)`  `= h/3`  
`:.h` ` = 3 xx sin(26.57°)= 1.34\ text{m}`  

Filed Under: Mechanics Tagged With: Band 4, smc-3718-30-Friction, smc-3718-50-Inclined planes

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)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`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%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-30-Friction, smc-3718-50-Inclined planes, smc-3718-60-Work Energy Power

Copyright © 2014–2025 SmarterEd.com.au · Log in