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Mechanics, SPEC2 2020 VCAA 20 MC

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

  1. `text(5.2 ms)^(−2)` upwards.
  2. `text(5.2 ms)^(−2)` downwards.
  3. `text(9.8 ms)^(−2)` downwards.
  4. `text(10.4 ms)^(−2)` upwards.
  5. `text(10.4 ms)^(−2)` downwards.
Show Answers Only

`A`

Show Worked Solution

`text(Assuming up is positive:)`

♦ Mean mark 43%.
`ma` `= 30 – 2g`
`2a` `= 10.4`
`:.a` `= 5.2\ text(ms)^(−2)`

 
`text{(i.e. acceleration is against the direction of travel.)}`

`=>A`

Mechanics, SPEC2-NHT 2019 VCAA 15 MC

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

  1.  30
  2.  45
  3.  60
  4.  75
  5.  90
Show Answers Only

`C`

Show Worked Solution

`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`

Mechanics, SPEC2 2013 VCAA 20 MC

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`

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ R\ text(be the reaction force of the floor on the 5kg parcel.)`

`sum F` `= 5g – R`
`5xx3` `= 5g-R`
`:. R` `=-15 +5g`

 
`=> A`

Mechanics, SPEC2-NHT 2017 VCAA 16 MC

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)`

Show Answers Only

`E`

Show Worked Solution

`sum F` `= (M+m) g – R`
`(M+m) a` `=(M+m)g-R`
`R` `= (M+m) (g-a)`

 
`=>E`

Mechanics, SPEC1 2015 VCAA 2

A 20 kg parcel sits on the floor of a lift.

  1. The lift is accelerating upwards at 1.2 ms¯².

     

    Find the reaction force of the lift floor on the parcel in newtons.  (2 marks)

  2. Find the acceleration of the lift downwards in ms¯² so that the reaction of the lift floor on the parcel is 166 N.  (2 marks)
Show Answers Only
  1. `220`
  2. `1.5\ text(ms)^-2`
Show Worked Solution
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)`

Mechanics, SPEC2-NHT 2018 VCAA 15 MC

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

Show Answers Only

`A`

Show Worked Solution

`sum F` `=80 text(g) – R`
`80 text(g) – R` `= 80 xx 1.2`
`:. R` `= 80xx9.8 – 80 xx 1.2`
  `= 688`

 
`=>  A`