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?
- 5
- 13
- 20
- 125
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?
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` ?
- `Mg`
- `Mg\ costheta`
- `Mg\ sintheta`
- `Mg\ tantheta`
Show Worked Solution
`=>B`
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?
- `P` and `W`
- `P` and `N`
- `F` and `W`
- `F` and `N`
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.
- Draw a free-body diagram that indicates the forces acting on the sled. Label the diagram. (2 marks)
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- Calculate the tension in the rope immediately before the point at which the sled and child begin to move. (3 marks)
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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%.
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` ?
- `N=W cos theta - P sin phi`
- `N=W cos theta + P sin phi`
- `N=W sin theta - P cos phi`
- `N=W sin theta + P cos phi`
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%.
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)
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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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