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Mechanics, SPEC2 2021 VCAA 15 MC

The diagram below shows a stationary body being acted on by four forces whose magnitudes are in newtons. The force of magnitude `F_1` newtons acts in the opposite direction to the force of magnitude 8 N.
 

The value of `F_1` is

  1. `8-2sqrt3`
  2. `2sqrt3`
  3. `8`
  4. `8 + 2sqrt3`
  5. `8-3sqrt3`
Show Answers Only

`A`

Show Worked Solution

`text(Resolve forces vertically:)`

`F_2sin60^@` `= 6sin30^@`
`F_2 sqrt3/2` `= 6 xx 1/2`
`F_2` `= 6/sqrt3`
  `= 2sqrt3`

 
`text(Resolve forces horizontally:)`

`F_1 + 6cos30^@` `= 2sqrt3 cos60^@ + 8`
`F_1 + 3sqrt3` `= sqrt3 + 8`
`F_1` `= 8 – 2sqrt3`

`=>\ A`

Mechanics, SPEC1 2020 VCAA 1

A 2 kg mass is initially at rest on a smooth horizontal surface. The mass is then acted on by two constant forces that cause the mass to move horizontally. One force has magnitude 10 N and acts in a direction 60° upwards from the horizontal, and the other force has magnitude 5 N and acts in a direction 30° upwards from the horizontal, as shown in the diagram below.
 


 

  1. Find the normal reaction force, in newtons, that the surface exerts on the mass.  (2 marks)
  2. Find the acceleration of the mass, in ms−2, after it begins to move.  (2 marks)
  3. Find how far the mass travels, in metres, during the first four seconds of motion.   (1 mark)
Show Answers Only
  1. `R = 2g – 5/2 – 5 sqrt 3\ text(N)`
  2. `a = 5/2 – (5 sqrt 3)/2\ text(ms)^(-2)`
  3. `20 – 10 sqrt 3\ text(m)`
Show Worked Solution
a.  

`text(Resolving forces vertically:)`

`2g` `= 5 sin 30 + 10 sin 60 + R`
`2g` `= 5/2 + 5 sqrt 3 + R`
`R` `= 2g – 5/2 – 5 sqrt 3\ text(N)`

 

b.    `2a` `= 10 cos 60 – 5 cos 30`
  `2a` `= 5 – (5 sqrt 3)/2`
  `:.a` `= 5/2 – (5 sqrt 3)/4\ text(ms)^(-2)`
Mean mark part (c) 51%.

 

c.    `text(Distance)` `= ut + 1/2 at^2`
    `= 0+ 1/2 (5/2 – (5 sqrt 3)/4) × 4^2`
    `= 20 – 10 sqrt 3\ text(m)`

Mechanics, SPEC2 2012 VCAA 14 MC

A particle is acted on by two forces, one of 6 newtons acting due south, the other of 4 newtons acting in the direction N60° W.

The magnitude of the resultant force, in newtons, acting on the particle is

  1. `10`
  2. `2sqrt7`
  3. `2sqrt19`
  4. `sqrt(52 - 24sqrt3)`
  5. `sqrt(52 + 24sqrt3)`
Show Answers Only

`B`

Show Worked Solution

Mean mark 51%.

`x^2` `= 6^2 + 4^2 – 2(6)(4) cos 60^@`
  `= 28`
   
`:. x` `= sqrt 28`
  `= 2 sqrt 7`

 
`=> B`

Mechanics, SPEC2 2014 VCAA 18 MC

A body on a horizontal smooth plane is acted upon by four forces, `underset ~F_1`, `underset ~F_2`, `underset ~F_3` and `underset ~F_4` as shown.

The force `underset ~F_1` acts in a northerly direction and the force `underset ~F_4` acts in a westerly direction.
 

 

Given that  `|\ underset ~F_1\ | = 1`, `|\ underset ~F_2\ | = 2`, `|\ underset ~F_3\ | = 4` and `|\ underset ~F_4\ | = 5`, the motion of the body is such that it

A.   is in equilibrium.

B.   moves to the west.

C.   moves to the north.

D.   moves in the direction 30° south of west.

E.   moves to the east.

Show Answers Only

`E`

Show Worked Solution

`text(Resolving forces horizontally:)`

♦ Mean mark 45%.

`sum F` `= underset ~F_2 cos30 + underset ~F_3 sin60 –  underset ~F_4`
  `= 2 xx sqrt3/2 + 4 xx sqrt3/2 -5`
  `~~0.2\ \ text(East)`

 
`text(Resolving forces vertically:)`

`sum F` `= underset ~F_1 + underset ~F_2 sin30 – underset ~F_3 cos 60`
  `= 1 + 2 xx 1/2 -4 xx 1/2`
  `= 0`

 
`:.\ text(Body moves to the east.)`

`=> E`

Mechanics, SPEC2 2017 VCAA 17 MC

Forces of 10 N and 8 N act on a body as shown below.
 


 

The resultant force acting on the body will, correct to one decimal place, have

  1. magnitude 15.6 N and act at 26.3° to the 10 N force.
  2. magnitude 9.2 N and act at 49.1° to the 10 N force.
  3. magnitude 15.6 N and act at 33.7° to the 10 N force.
  4. magnitude 9.2 N and act at 70.9° to the 10 N force.
  5. magnitude 15.6 N and act at 49.1° to the 10 N force.
Show Answers Only

`A`

Show Worked Solution

 
 

`|∑underset~F|^2` `= 10^2 + 8^2 – 2 xx 10 xx 8cos120°`
`|∑underset~F|` `= sqrt244`
  `~~ 15.6`

 

`(sin(theta))/8` `= (sin(120°))/15.62`
`sin(theta)` `= (4sqrt3)/15.62`
`:. theta` `~~ 26.3°`

 
`=> A`

Mechanics, SPEC2 2017 VCAA 16 MC

An object of mass 20 kg, initially at rest, is pulled along a rough horizontal surface by a force of 80 N acting at an angle of 40° upwards from the horizontal. A friction force of 20 N opposes the motion.

After the pulling force has acted for 5 seconds, the magnitude of the momentum, in kg ms−1, of the object is closest to

  1.     10
  2.     40
  3.   160
  4.   210
  5. 4100
Show Answers Only

`D`

Show Worked Solution


 

`∑underset~F = 80cos(40°) – 20`

`ma` `= 80cos(40°) – 20`
`a` `=4 cos(40°) – 1`

 

`Deltav` `= 0+at`
  `= 5(4cos(40°) – 1)`

 

`m Deltav` `= 100(4cos(40°) – 1)`
  `~~ 206.4`

 
`=> D`