Luggage at an airport is delivered to its owners via a ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.
- By resolving forces parallel to the ramp, show that the magnitude of the acceleration, `ddot{x}\ text(ms)^(-2)`, of the suitcase down the ramp is given by `ddot{x} = (g-2v)/2`. (2 marks)
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- Using 9.8 `text(ms)^(-2)` as the acceleration due to gravity, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form `x = bv + c\ log_e(c/(c-v))`, where `b, c in R`. (3 marks)
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Show Worked Solution
a.
| `sumF` | `=20g sin 30^@-20v` | |
| `m ddot{x}` | `=10g-20v` | |
| `ddot{x}` | `= (g-2v)/2` |
b. `text{Using}\ \ ddot{x}=v *(dv)/(dx):`
| `(dv)/(dx)` | `= (g-2v)/(2v)` |
| `(dx)/(dv)` | `= (2v)/(g-2v)` |
| `(dx)/(dv)` | `= -(2v)/(2v-g)=-((2v-g + g))/(2v-g)= -1-g/(2v-g)` |
`text{Find the distance travelled:}`
| `x` | `= int_0^v-1-g/(2v-g)\ dv` |
| `= int_0^v-1-g/2 (2/(2v-g))\ dv` | |
| `= [-v-4.9 xx ln\ |2v-g|]_0^v` |
`text(When)\ \ x=0, v=0:`
`2v-g < 0\ \ =>\ \ |2v-g| = g-2v`
| `x` | `= [-v-4.9 ln (g-2v)]_0^v` |
| `= -v-4.9 ln (g-2v)-(0-4.9 ln (g))` | |
| `= -v + 4.9 ln (g/(g-2v))` | |
| `=-v + 4.9 ln(9.8/(9.8-2v))` | |
| `= -v + 4.9 ln ((4.9)/(4.9-v))` |