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A body of mass 5 kg is acted on by a net force of magnitude `F` newtons. This force causes the body to move so that its velocity, `v\ text(ms)^(-1)`, along a straight line of motion is given by `v = 3 + 2x`, where `x` metres is the position of the body at time `t` seconds.
When `x = 2, F` is equal to
`D`
| `a` | `= v · (dv)/(dx)` |
| `= (3 + 2x) · 2` |
`text(Find)\ F\ text(when)\ x = 2:`
| `F` | `= ma` |
| `= 5 xx 2(3 + 2 xx 2)` | |
| `= 70\ text(N)` |
`=>\ D`
A particle of mass 3 kg is acted on by a variable force, so that its velocity `v` m/s when the particle is `x` m from the origin is given by `v = x^2`.
The force acting on the particle when `x = 2`, in newtons, is
A. 4
B. 12
C. 16
D. 36
E. 48
`E`
| `v` | `=x^2` |
| `a` | `= v *(dv)/(dx)` |
| `= x^2 xx 2x` | |
| `= 2x^3` |
♦ Mean mark 50%.
`text(When)\ \ x=2:`
| `a` | `= 2 xx 2^3` |
| `=16` | |
| `:.F` | `= 3 xx 16` |
| `=48` |
`=> E`
A constant force of magnitude `F` newtons accelerates a particle of mass 2 kg in a straight line from rest to 12 ms`\ ^(−1)` over a distance of 16 m.
It follows that
`B`
`u = 0, \ v = 12, \ s = 16`
| `v^2` | `= u^2 + 2as` |
| `144` | `= 0 + 32a` |
| `a` | `= 9/2\ \ \ text{(by CAS)}` |
| `:. F` | `= 2(9/2)` |
| `= 9` |
`=> B`
A particle of mass 2 kg moves in a straight line with an initial velocity of 20 m/s. A constant force opposing the direction of the motion acts on the particle so that after 4 seconds its velocity is 2 m/s.
The magnitude of the force, in newtons, is
A. 4.5
B. 6
C. 9
D. 18
E. 36
`C`
`u = 20, \ t = 4, \ v = 2`
`v = u + at`
| `v` | `=u + at` |
| `2` | `= 20 + 4a` |
| `a` | `= −9/2` |
| `|underset~F|` | `= 2 xx |−9/2|=9` |
`=> C`
A variable force of `F` newtons acts on a 3 kg mass so that it moves in a straight line. At time `t` seconds, `t >= 0`, its velocity `v` metres per second and position `x` metres from the origin are given by `v = 3 - x^2`.
It follows that
A. `F = -2x`
B. `F = -6x`
C. `F = 2x^3 - 6x`
D. `F = 6x^3 - 18x`
E. `F = 9x - 3x^3`
`D`
`(dv)/(dx) = -2x`
`a = v (dv)/(dx) = -2x(3 – x^2)`
| `:. F` | `=ma` |
| `= -6x(3 – x^2)` | |
| `= 6x^3-18x` |
`=> D`
A 3 kg mass has velocity `v\ text(ms)^(-1)`, where `v = 2 arctan sqrt x` when it has a displacement `x` metres from the origin, `x > 0`.
Find the net force, `F` newtons, acting on the mass in terms of `x`. (3 marks)
`F = (6 tan^(-1)(sqrt x))/(sqrt x(1 + x))`
| `v` | `= 2 tan^(-1) (x^(1/2))` |
| `(dv)/(dx)` | `= 1/2 xx x^(-1/2) xx 2/(1 + (sqrt x)^2)` |
| `= 1/(sqrt x (1 + x))` |
| `a=v*(dv)/(dx)` | `= 2 tan^(-1) (sqrt x) xx 1/(sqrt x (1 + x))` |
| `= {2 tan^(-1) (sqrt x)}/{sqrt x (1 + x)}` |
| `:. F` | `= ma` |
| `= 3a` | |
| `= (6 tan^(-1)(sqrt x))/(sqrt x(1 + x))` |
A constant force of magnitude `P` newtons accelerates a particle of mass 8 kg in a straight line from a speed of 4 ms¯¹ to a speed of 20 ms¯¹ over a distance of 15 m.
The magnitude of `P` is
A. 9.8
B. 12.5
C. 12.8
D. 100
E. 102.4
`E`
`u = 4, quad v = 20, quad s = 15`
| `v^2` | `= u^2 + 2as` |
| `20^2` | `= 4^2 + 30a` |
| `400` | `= 16 + 30a` |
| `384` | `= 30a` |
| `a` | `= 64/5` |
| `:. P` | `= 8a` |
| `= (8 xx 64)/5` | |
| `= 102.4` |
`=> E`