MECHANICS: FORCE AND NEWTON'S LAWS - QUESTIONS AND ANSWERS

More in this category: « MECHANICS: FORCE AND NEWTON'S LAWS - PHYSICAL SCIENCES PAPER 1 STUDY GUIDES AND NOTES MOMENTUM AND IMPULSE - PHYSICAL SCIENCES PAPER 1 STUDY GUIDES AND NOTES »
Share via Whatsapp Join our WhatsApp Group Join our Telegram Group

Worked example 1
1

  • Just as one vector is the sum of two vectors, we can also find two vectors to make up one vector.
  • In mechanics, it is often useful to break up a vector into two component vectors, one horizontal and the other vertical. We use basic trigonometry to find the components.

Worked example 2
Example: Vector R makes an angle θ with the x-axis. R is broken into component vectors Rx and Ry.
2
cos θ = adj = R x
            hyp    R
∴ Rcos θ = Rx
sin θ = opp = R y
            hyp    R
∴ Rsin θ = Ry

Rx = a cos θ and Ry = sin Ry, where Ry is the angle between a and the x-axis (Rx).

Worked example 3
If vector R has a magnitude of 5 and is at an angle of θ = 36,86°, the components are Fx = 5 cos 36,86° = +4 and Ry = 5 sin 36,86° = +3.

Worked example 4
Draw a force diagram and a free body diagram for an object hanging from a rope or a cable.
9

Solution
10

Worked example 5
If an object with a mass of 40 kg slides down a surface which has a coefficient of kinetic friction μk = 0,14, and a slope of 15°, what is the net force on the object as it slides down the surface? Use the diagram to help you.
12

Solution
Weight component down the slope
=Fg sin θ = mg sin 15° = 40 × 9,8 × 0,26 = 101,92 N.
Frictional force up the slope
Fg·μk cos θ = 40 × 9,8 × 0,14 × 0,96 = 52,68 N
∴ Net force down the slope is: 


Worked example 6
Calculate the components of the weight of an object when it is resting on a surface which slopes at an angle of θ.

Solution
θ + a = 90° and a + c = 90°. So c = θ.
The component of the weight perpendicular to the surface with a slope of θ is
Fg⊥ =Fg cos θ and parallel to the surface is
Fg|| =Fg sin θ. 
13

Worked example 7
John exerts a force of 100 N to the right on a box resting on a rough, horizontal surface. Sarah exerts a force of 50 N to the left on the box. The friction between the box and the surface is 5 N. Draw a force diagram and calculate the resultant force acting on the box.
14

Solution
Fnet = ΣF = (+FJohn) + (−FSarah) + (−Ff)
=FJohn –FSarah –Ff
= 100 – 50 – 5 = 45 N ∴Fnet= 45 N to the right

Activity 1
A box of mass 100 kg slides down a rough slope which forms an angle of 30° to the horizontal. The friction that acts on the box is 20 N. Draw a free body diagram representing all the forces acting on the object and calculate the resultant force acting on the box and causing it to slide. Perpendicular forces may be ignored. (12)

Solution
Let the direction down the slope be positive.
Then Ff = –20N
Fnet = ΣF = (+Fgll) + (Ff ) 
= Fgll + Ff
= mg·sin α + Ff 
= (100)(9,8)(sin 30°)+( – 20)
= 490 – 20 = 470 N
∴ 470 N, down the slope
15

[12]

Activity 2
R and S are two positively charged spheres. P is a negatively charged sphere. Sphere R exerts an electrostatic force of 0,2 N on P and sphere S exerts a force of 0,6 N on sphere P.
rps
Draw a free body diagram for sphere P and then calculate the resultant force on sphere P. (5)

Solution
16
F net = ΣF = (FS on P) + (FR on P) 
= 0,6 + (–0,2)
= 0,4 N ∴ 0,4 N towards S [5]
Note:FR = –0,2 N 

Remember: Opposite charges attract. Therefore R attracts P and S attracts P.

Activity 3
Three identical spheres X, Y and Z are in the same horizontal plane.
Spheres X and Z are both positive and sphere Y is negative. Sphere Y exerts an electrostatic force of 450 N on sphere X and sphere Z exerts an electrostatic force of 350 N on sphere X.
x

  1. Draw a free body diagram for sphere X and indicate the electrostatic forces acting on it. (2)
  2. Calculate the magnitude of the resultant electrostatic force on sphere X. (8)
    [10]

Solutions

  1. Free body diagram:
    17
  2. F2net = F2Z on X + F2Y on X (pythagoras)
    = 3502 + 4502 = 325 000
    = ∴Fnet = √325 000 = 570,09 N
    18
    [10] 

Worked example 8
The diagram shows the forces on a trolley moving with constant velocity.
5

  1. A man pushes a loaded trolley with constant velocity along a horizontal floor. The trolley and load have a mass of 56 kg and the friction of the moving trolley is 2,1 N. Calculate the force the man exerts to push the trolley along the floor.
  2. If he then pushes the trolley with a force of 2,5 N to the right, calculate the acceleration of the trolley.

Solutions

  1. The acceleration = 0, so the net force is equal to zero. The force the person pushes with is equal and opposite to the force of kinetic friction on the trolley.
    Fpush = Ftrolley = 2,1 N
  2. The diagram shows the free body diagram.
    The net force is 0,4 N in the forward direction.
    The trolley accelerates forward:
    a = Fnet = 0,4 = 7,14 × 10–3 m·s–2
           m       56
    forward / to the right. 
    19

Worked example 9
A resultant force F is applied to an object of mass m and the object accelerates at a .
What will the object’s acceleration be if the resultant force acting on the object is tripled?

Solution
m is constant ∴ a ∝ F and if the force is tripled (from F to 3F ), the acceleration will also triple ∴ the object will accelerate at 3a . 

Worked example 10
A constant resultant force
F is applied to objects of masses m and 2 m. If the object of mass m accelerates at a , what will the acceleration of the other object be?

Solution
F is constant ∴ a ∝ 1/m If the mass doubles (from m to 2 m), the acceleration will halve
∴ the object of mass 2 m accelerates at ½ a . 

Activity 4
Two boxes, A and B, are lying on a table and are connected by a piece of string. The mass of box A is 3 kg and the mass of box B is 2 kg. Assume that the mass of the string is very small, so we can ignore it. A 30 N pulling force, pointing to the right, is applied to box B, causing the two boxes to move. The surface acts with a frictional force of 5,9 N on box A and 4,1 N on box B.
22

  1. Calculate the acceleration of boxes A and B. (14)
  2. Calculate the magnitude of the tension on the string. (5)
    [19]

Solution

  1. We are going to take the whole system as a unit.
    23
    Data:
    mA = 3 kg, mB = 2 kg 
    mB = 2 kg mT = mA + mB = 3 kg + 2 kg = 5 kg
    FA = 30 N to the right
    FfA = 5,9 N to the left
    FfB = 4,1 N to the left
    FfT = FfA + FfB = 5,9 + 4,1 = 10 N to the left
    a = ? to the right
    T = ? (9)
    FRTx = mTa (from F = ma) 
    FA + FfT = mTa
    (30) + (–10) = 5a
    a = 4 m/s2 to the right  (5)
  2. To calculate the tension you may use box A or box B
    Tension using box A
    24
    FRAx = mAa 
    T + FfA = mAa 
    T – 5,9 = (3)(4) 
    T = 12 + 5,9
    T = 17,9 N  (5)
    Tension using box B
    25
    FRBx = mBa
    FA + T + FfB = mBa 
    30 – T – 4,1 = (2)(4) 
    –T + 25,9 = 8
    –T = –17,9
    T = 17,9 N 
    [19] 


Activity 5
The sketch below shows a block of 8,5 kg at equilibrium on an inclined (sloping) plane (surface).
26
Calculate:

  1. The magnitude of the tension in the cord. (12)
  2. The magnitude of the normal force acting on the block. (6)
  3. The magnitude of the block’s acceleration, If the cord is cut. (4)
    [22]

Solution

  1. Data
    m = 8,5 kg ; α = 30º ; vi = 0 ; ai = 0
    Let’s make the free body diagram of forces. (6)
    27
    The gravitational force is not in the direction of any axis then we have to determine its components on the x-axis and y-axis.
    Applying Newton’s First Law
    ΣF = 0
    Fg + T = 0 
    Working with the projections of the forces on the x-axis we get:
    Fgx – T = 0
    Fg·sin α – T = 0 
    m·g·sin 30° – T = 0 
    8,5 × 9,8 × 0,5 – T = 0
    41,65 – T = 0
    T = 41,65 N (6)
  2. Working on the y-axis
    N – Fgy = 0
    N – (Fg·cos α) = 0 
    N – (m·g·cos 30º) = 0
    N – (8,5 × 9,8 × 0,866 ) = 0
    N – 73,1 = 0
    N = 73,1 N  (6)
  3. Applying Newton’s Second Law.
    ΣFx= max 
    If the cord is cut there is no tension force acting on the block and there is only one force acting on the direction of the x-axis, causing acceleration to the block.
    Working with the projections
    Fgx = m·a
    Fg·sin α = m·a
    mg·sin 30° = m·a 
    Simplifying:
    g·sin 30° = a
    a = g·sin 30°
    a = 9,8 × 0,5 
    a = 4,9 m·s–2  (4)
    [22] 


Activity 6
Two blocks of 25 kg and 15 kg are connected by a light string on a horizontal surface. Assume that the string cannot stretch. A force of magnitude 240 N is applied to the block of 15 kg forming an angle of 60° with the horizontal as shown in the sketch below. The coefficient of kinetic friction is 0,20.
28

  1. State Newton’s Second Law of Motion in words. (7)
  2. Draw a free body diagram for each block. (8)
  3. Calculate the magnitude of the acceleration of the blocks. (14)
    [29]

Solutions

  1. If a resultant force  acts on a body, it will cause the body to accelerate in the direction of the resultant  force. The acceleration of the body will be directly proportional to the resultant force and inversely 3 proportional to the mass  of the body. (7)
  2. 29(8)
  3. ption 1
    Taking the objects as a system
    FRx = ma 
    Fx + FfT = max
    Fx – FfT = mTax
    Fx – (Ff1 + Ff1) = (m1 + m2)ax
    Fx – (μN1 + μN2) = (m1 + m2)ax
    Note: In this series of solutions we have omitted (taken out) the vector arrow above F and a; this is just to make it easier to read the solution.
    We have to calculate the normal force for both blocks
    N1 = Fg = m1g
    N2 = m2g – F·sin 60˚ 
    F·cos 60˚ – [μm1g + μ(m2g – F·sin 60˚)] = (m1 + m2)ax
    (240·cos 60˚) – [(0,2)(25)(9,8) + (0,2)[(15)(9,8)] – 240·sin 60˚]]
    = (25+15)ax
    120 – [49 + (0,2)(147 – 207,85)] = 40ax
    83,17 = 40ax
    ax = 2,08 m/s2 
    Option 2
    Applying Newton’s Second Law of motion to each object individually
    FR = m1a
    For object 1:
    T = Ff1 = m1ax
    T – Ff1 = m1ax
    T – μm1g = m1ax
    For object 2:
    FR2x = m2a 
    Fx + T + Ff2 = m2ax
    Fx – T – Ff2 = m2ax
    F·cos 60˚ – T – μN1 = m1ax
    F·cos 60˚ – T – μ(m2g – F·sin 60˚) = m2ax
    Adding equation (1) and (2).
    T – μm1g + F·cos 60˚– T – μ(m2g –F·sin 60˚) = m1ax + m2ax
    Taking out T andax: – μm1g + F·cos 60˚– μ(m2g –F·sin 60˚) = (m1 + m2)ax
    [–(0,2)(25)(9,8)] + [240·cos 60˚– (0,2)[(15)(9,8) – (240·sin 60˚)] =
    (25 + 15)ax
    (–49 + 120) – (0,2)(147 – 207,85) = 40ax
    71 + 12,17 = 40ax
    83,17 = 40ax
    ax = 2,08 m/s2 (14)
    [29] 


Activity 7
The sketch below shows two blocks connected by a string of negligible mass that passes over a frictionless pulley also of negligible mass.
The arrangement is known as Atwood’s machine. One block has mass
m1 = 2 kg and the other has mass m2= 4 kg.
The blocks have just this instant been released from rest.

  1. Draw a free body diagram of all the forces acting on each block. (6)
  2. Calculate the magnitude of the acceleration of the system. (7)
  3. Calculate the magnitude of the tension in the string. (4)
  4. Compare the magnitude of the net force on m1 with the net force on m2. (1)
    Write down only GREATER THAN, SMALLER THAN or EQUAL TO.
  5. Will the pulley rotate clockwise or anticlockwise? (1)
    [19]

Solutions

  1. See diagram below:
    30
    Note: In this series of solutions we have omitted (taken out) the vector arrow above F and a; this is just to make it easier to read the solution.
    (6)
  2. ƩFnet = ma
    For the 2 kg block (+ upwards)
    T – Fg1 = m1a
    T –m1g = m1a 
    T – 2 × 9,8 = 2a 
    For the 4 kg block (+ downwards)
    –T + m2g = m2a
    –T+ 4 × 9,8 = m2a 
    Solving the system of equations
    T – 2 × 9,8 –T + 4 × 9,8 = (2 + 4)a
    2 × 9,8 = 6 a
    a = + 3,27 m·s–2 (upwards)
    a = 3,27 m·s–2  (7)
  3. Option 1
    T – (2m 1m 2) × g 
             m1m2
    T = (2×2×4) × 9,8
             2+4
    T = 26,13 N
    Option 2
    T – Fg1 = m1a OR T= m1(a+g)
    T = 2 (3,27 + 9,8) 
    T = 26,14 N 
    Option 3
    –T + m2g = m2a OR – T = m2a – m2g OR T = – m2a + m2g
    T = 4(–3,27 + 9,8)
    T = 26,12 N (4)
  4. Smaller than. (1)
  5. Anticlockwise. (1)
    [19] 

 

Activity 8
A 4 kg block on a horizontal, rough surface is connected to a 8 kg block by a light string that passes over a frictionless pulley as shown below. Assume that the string cannot stretch. The coefficient of kinetic (dynamic) friction between the block of 4 kg and the surface is 0,6.
31

  1. Draw a free body diagram of all the forces acting on both blocks. (6)
  2. Calculate the acceleration of the system. (10)
  3. Calculate the magnitude of the tension in the string. (3)
  4. Calculate the magnitude of the frictional force that acts on the 4 kg block. (4)
  5. Calculate the apparent weight of the 8 kg block. (4)
  6. How does the apparent weight of the 8 kg block compare with its true weight? Write down only GREATER THAN, EQUAL TO or LESS THAN. (1)
  7. How does the apparent weight of the 4 kg block compare with its true weight? Write down only GREATER THAN, EQUAL TO or LESS THAN. (1)
    [29]

Solutions

  1. 32(6)
    Let’s take the direction of motion as positive
  2. Let’s apply Newton’s Second Law of Motion to each block.
    ΣF = ma
    Block of 4 kg (A)
    In the x direction (horizontal)
    T – ff = mAa
    T – μN = mAa
    T – μmAg = mAa (call this Equation 1)
    Block of 4 kg (A)
    In y direction (up/down)
    N – Fg = 0 (not moving horizontally)
    N = Fg = mg
    Block of 8 kg (B)
    In the x direction (horizontal)
    –T + Fg = mBa
    –T + mBg = mBa (call this Equation 2)
    Solving the system of equations (adding Equation 1 and 2)
    T – μmAg – T + mBg = mAa + mBa
    Removing T and isolating a:
    –μmAg + mBg = (mA + mB)a
    –(0,6)(4)(9,8) + (8)(9,8) = (4 + 8)a
    54,88 = 12a
    a = 4,57 m/s2 (10)
  3. Using Equation 2
    –T + mBg = mBa
    –T = 8 × 4,57 – (8 × 9,8)
    T = 41,84 N (3)
    Note: In this series of solutions we have omitted (taken out) the vector arrow above F and a; this is just to make it easier to read the solution
    Using Equation 1
    T – μmAg = mAa
    T – (0,6)(4)(9,8) = (4)(4,57)
    T = (0,6)(4)(9,8) + (4)(4,57)
    T = 41,8 N (3)
  4. ff = μN
    N = mg
    ff = μmg
    ff = 0,6 × 4 × 9,8
    ff = 23,52 N (4)
  5. –T + mBg = mBa
    –T = –8 × 4,57 + (8 × 9,8)
    Apparent weight = T = 41,84 N (4)
  6. Less than 3 (1)
  7. Equal to 3 (1)
    [29]

Worked example 11
36
A force of gravitational attraction exists between the earth with mass m1 and a person with mass m2. The force on m1 is F1 and the force on m2 is F2.
Compare the magnitudes (sizes) of these forces and state the name of the law which explains your answer.

Solution
F1 = – F2 according to Newton’s Third Law of Motion:
The force between the earth (m1) and a person (m2) standing on its surface:
∴ m2g = G m 1m 2
                      r2
∴ g = G m1
             r2 

Worked example 12
The diagram shows a ball A of mass 0,01 kg which is 1 m (measured from centre to centre) from another ball B of mass 520 g. Calculate the magnitude of the force of ball A on ball B.
38

Solution
F = G m 1m 2 = 6.67 x 10-11 x 0.01 x 0.52
             r2                         12
= 3,57 × 10–14 N
It is important to understand and be able to compare mass and weight.

Worked example 13
An object weighs 720 N on earth. It orbits the Earth in a satellite at a height equal to the earth’s diameter, above the surface of the Earth. What does the object weigh on the satellite?
Hint: diameter = 2 × radius

Step by step
Step 1. Determine the number of radii from the centre of the Earth.
On the Earth’s surface, the object is 1 radius from the Earth’s centre.
Step 2. Determine how many times the distance between the object and the centre of the Earth has increased.
In orbit, the object is 1 diameter = 2 radii above the surface.
Step 3. Square this number (multiply it by itself).
So the object is 3 radii from the centre. (3)2 = 9
39
Step 4. The force has decreased this number of times because F ∝ 1 
                                                                                                           r2
∴ the gravitational force on the object has decreased 9 times because F ∝ 1 
                                                                                                                      r2

Step 5. Divide the value of the force (or weight) by the value calculated in step 3.
∴ its weight on the satellite is 720 N ÷ 9 = 80 N
Gravitational acceleration on planets other than earth
Newton’s universal law of gravitation can be used to calculate the acceleration due to the force of gravity on any planet.
If the mass and radius of a planet are known, we can calculate g for that planet.

Worked example 14
The Mars Rover is an automated vehicle that has been sent to explore the surface of the planet Mars.
If the value of acceleration due to gravity on the planet Mars is g Mars = 3,7 m·s–2.
Calculate the weight of the Mars rover on Mars if it has a mass of 174 kg.

Solution
w Mars = g Mars × mobject = 3,7 × 174 = 643,8 N towards the centre of the planet Mars.

Last modified on Tuesday, 21 September 2021 07:37
Published in Physical Sciences Paper 1 Grade 12 Study Guide and Notes

Related items

More in this category: « MECHANICS: FORCE AND NEWTON'S LAWS - PHYSICAL SCIENCES PAPER 1 STUDY GUIDES AND NOTES MOMENTUM AND IMPULSE - PHYSICAL SCIENCES PAPER 1 STUDY GUIDES AND NOTES »
back to top