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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
An object of mass 2 kg is suspended from a spring balance that is inside a lift travelling downwards.
If the reading on the spring balance is 30 N, the acceleration of the lift is
`text(Assuming up is positive:)`
| `ma` | `= 30 – 2g` |
| `2a` | `= 10.4` |
| `:.a` | `= 5.2\ text(ms)^(−2)` |
`text{(i.e. acceleration is against the direction of travel.)}`
`=>A`
A lift accelerates from rest at a constant rate until it reaches a speed of 3 ms−1. It continues at this speed for 10 seconds and then decelerates at a constant rate before coming to rest. The total travel time for the lift is 30 seconds.
The total distance, in metres, travelled by the lift is
`text(Consider the velocity graph:)`
`t_1 -> t_2 = text(10 seconds)`
`text(Distance = area of trapezium)`
`= 1/2 xx 3(10 + 30)`
`= 60\ text(m)`
`=>\ C`
A ball is thrown vertically upwards with an initial velocity of `7sqrt6` ms−1, and is subject to gravity and air resistance. The acceleration of the ball is given by `overset(¨)x = −(9.8 + 0.1v^2)`, where `v` ms−1 is its velocity when it is at a height of `x` metres above ground level.
The maximum height, in metres, reached by the ball is
An object of mass 2 kg is travelling horizontally in a straight line at a constant velocity of magnitude 2 ms−1. The object is hit in such a way that it deflects 30° from its original path, continuing at the same speed in a straight line.
The magnitude, correct to two decimal places, of the change of momentum, in kg ms−1, of the object is
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
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
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
A 5 kg parcel is on the floor of a lift that is accelerating downwards at 3 m/s².
The reaction, in newtons, of the floor of the lift on the parcel is
A. `−15 + 5g`
B. `15 + 5g`
C. `−15 + 3g`
D. `−15 − 5g`
E. `15 + 3g`
A body of mass 3 kg is moving to the left in a straight line at 2 `text(ms)^(-1)`. It experiences a force for a period of time, after which it is then moving to the right at 2 `text(ms)^(-1)`.
The change in momentum of the particle, in kg `text(ms)^(-1)`, in the direction of the final motion is
A. − 6
B. 0
C. 4
D. 6
E. 12
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`
A person of mass `M` kg carrying a bag of mass `m` kg is standing in a lift that is accelerating downwards at `a\ text(ms)^(-2)`.
The force of the lift floor acting on the person has magnitude
A. `Mg + mg`
B. `Mg + (M + m)a`
C. `Mg - (M + m)a`
D. `(M + m) (g + a)`
E. `(M + m) (g - a)`
| `sum F` | `= (M+m) g – R` |
| `(M+m) a` | `=(M+m)g-R` |
| `R` | `= (M+m) (g-a)` |
`=>E`
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)
A 20 kg parcel sits on the floor of a lift.
Find the reaction force of the lift floor on the parcel in newtons. (2 marks)
| a. |  |
`text(Acceleration:)\ \ ↑ 1.2\ text(ms)^(−2)`
| `∑F` | `= 20 xx 1.2` |
| `= R – 20text(g)` |
| `R – 20 xx 9.8` | `=20 xx 1.2` |
| `:. R` | `= 20(1.2 + 9.8)` |
| `= 20 xx 11` | |
| `= 220\ text(N upwards)` |
b. `↓a\ text(ms)^(−2)`
| `∑F` | `= 20a` |
| `= 20text(g) – R_2` |
| `20a` | `= 20text(g) – 166` |
| `= 20text(g) – 2 xx 83` | |
| `= 20text(g) – 20 xx 8.3` | |
| `:. a` | `= text(g) – 8.3` |
| `= 9.8 – 8.3` | |
| `= 1.5\ text(ms)^(−2)` |
A body of mass 2 kg is moving in a straight line with constant velocity when an external force of `8 N` is applied in the direction of motion for `t` seconds.
If the body experiences a change in momentum of 40 kg ms¯¹, then `t` is
A. 3
B. 4
C. 5
D. 6
E. 7
An 80 kg person stands in an elevator that is accelerating downwards at `1.2\ text(ms)^(-2)`.
The reaction force of the elevator floor on the person, in newtons, is
A. 688
B. 704
C. 784
D. 880
E. 896
| `sum F` | `=80 text(g) – R` |
| `80 text(g) – R` | `= 80 xx 1.2` |
| `:. R` | `= 80xx9.8 – 80 xx 1.2` |
| `= 688` |
`=> A`
A tourist standing in the basket of a hot air balloon is ascending at 2 ms¯¹. The tourist drops a camera over the side when the balloon is 50 m above the ground.
Neglecting air resistance, the time in seconds, correct to the nearest tenth of a second, taken for the camera to hit the ground is
A. 2.3
B. 2.4
C. 3.0
D. 3.2
E. 3.4
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