A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200 newtons at an angle of 30° to the horizontal. The box experiences a resistive force of `0.3R` newtons, where `R` is the normal force, as shown in the diagram.
Take the acceleration `g` due to gravity to be 10m/s2.
- By resolving the forces vertically, show that `R =400`. (2 marks)
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- Show that the net force horizontally is approximately 53.2 newtons. (2 marks)
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- Find the velocity of the box after the first three seconds. (2 marks)
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Show Worked Solution
i.
`text{Resolving forces vertically:}`
| `R + 200 \ sin 30^@` | `= 50g` |
| `R + 200 xx frac{1}{2}` | `= 50 xx 10` |
| `R + 100` | `= 500` |
| `therefore \ R` | `= 400 \ text(N)` |
ii. `text{Resolving forces horizontally:}`
| `text{Net Force}` | `= 200 \ cos 30^@ – 0.3 R` |
| `= 200 xx frac{sqrt3}{2} – 0.3 xx 400` | |
| `= 100 sqrt3 – 120` | |
| `= 53.2 \ text{N (to 1 d. p.)}` |
| iii. | `F` | `=ma` |
| `50 a` | `=100 sqrt300 – 120` | |
| `a` | `= frac{100 sqrt3 – 120}{50}\ text(ms)^(-2)` |
`text{Initially} \ \ u = 0,`
| `v` | `= u + at` |
| `v_(t=3)` | `= 0 + frac{100 sqrt3 – 120}{50} xx 3` |
| `= 3.1923 \ …` | |
| `= 3.19 \ text{ms}^-1 \ text{(to 2 d.p.)}` |