Target: 10 Questions in 10 minutes

An IB Physics data booklet is helpful

Q1-3: These questions are about the forces acting on a meteor as it falls to Earth.

 

 1. When drawing an arrow to show the size of the force of gravity on the meteor, which of these images shows the correct placement for the arrow?

force arrows on a meteor

2. The force of gravity as shown by the arrow in Q1 can be labelled in many ways. Which of these is likely to marked as incorrect?

  • A. weight
  • B. mg
  • C. gravity
  • D. downwards force of gravity from the Earth

3. Which of these diagrams correctly shows the size of the forces on the meteor as it is moving at a very high velocity downwards but slowing down due to air resistance? The meteor has been drawn as a point.

force arrows on 4 asteroids #2

Q4-6. The diagram below shows a stationary block on a slope. The mass of the block is m.

4. Which of the diagrams below correctly shows all the forces acting on the block?

 

block on slope

5. In which direction is the resultant force acting in question 4?

  • A. Up the slope.
  • B. Down the slope.
  • C. Towards the centre of the Earth.
  • D. There is no resultant force.

6. Which of these statements is true?

  • A. Normal force = m.g.sinθ
  • B. Normal force= m.g.cosθ
  • C. Normal force= m.sinθ
  • D. Normal force= m.cosθ

7. A stone of mass 0.5 kg is dropped from a height. At one point in the drop the stone is accelerating at 3 ms-2. At this point, the force from air resistance is approximately..

  • A. 5 N
  • B. 3.5 N
  • C. 3 N
  • D. 1.5 N

 

8&9. These questions are about a mass on a spring of spring constant k = 500 N m-1.
(For all spring questions, assume Hooke's law is obeyed.)

 

mass spring system
Svjo - wikimedia commons - edited CC3.0

8. What force is applied to give an extension of 5 cm?

  • A. 100 N
  • B. 10 N
  • C. 2500 N
  • D. 25 N

9. A mass of 4 kg as added to the spring. The extension will be:

  • A. 1.2 cm
  • B. 12 cm
  • C. 8 cm
  • D. 0.08 cm

10. A different spring has a length of 15 cm with a mass of 2 kg added, and 19 cm with a mass of 3 kg. What is the spring constant k for this spring, and what is the length L when 5 kg is added?

 
k / N m-1 L / cm
A
2.5 20
B
2.5 27
C
250 20
D
250 27

 

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Question 1:

1. Correct Placement for the Force of Gravity

The force of gravity acts on the center of mass of an object and is always directed towards the center of the larger celestial body. For the meteor, the force of gravity is a pull towards the center of the Earth, downwards.


The diagram that shows a pulling force acting on the centre of mass is diagram A.

(Edited - the Silverback)


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Question 2:

Incorrect Label for the Force of Gravity

The force of gravity on an object is commonly called weight. The mathematical expression for weight is mg, where m is the mass of the object and g is the acceleration due to gravity. When an object is falling, the force is often described as the downwards force of gravity from the Earth.

However, the term "gravity" is not a force in itself. It is the fundamental interaction or phenomenon that causes the force. The force itself is the gravitational force or, more commonly, weight. Therefore, using "gravity" as a label for the force is a common mistake and is often considered incorrect in physics.

This means answer C is correct.

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Question 3:

Let's break down the components of this diagram:

The Meteor: Represent it as a small circle or dot (O) to signify it's a point.

Force of Gravity (Weight):

Direction: Draw an arrow pointing directly downwards from the meteor. This arrow should point towards the center of the Earth.

Label: Label this arrow "Force of Gravity," "Weight," or "mg."

Size: The length of this arrow represents the magnitude of the gravitational force.

Air Resistance (Drag Force):

Direction: Draw an arrow pointing directly upwards from the meteor, opposing the direction of motion.

Label: Label this arrow "Air Resistance" or "Drag Force."

Size: This is the crucial part for showing the meteor is slowing down. Since the meteor is decelerating (slowing down), the upward force of air resistance must be larger than the downward force of gravity. Therefore, draw this arrow significantly longer than the gravity arrow.

Key things this diagram conveys:

Net Force: Because the upward air resistance arrow is longer than the downward gravity arrow, the net force on the meteor is upwards.

Deceleration: An upward net force on an object moving downwards means it is decelerating, or slowing down, which matches the problem description.

Point Object: The forces are drawn originating from or acting upon the "point" of the meteor.

Here's an illustrative diagram:

^
|
| Air Resistance (F_air) - LONGER arrow
|
|
O <-- Meteor (point)
|
| Force of Gravity (mg / Weight) - SHORTER arrow
v

This corresponds to diagram B.


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Question 4:

The correct diagram showing all the forces acting on a stationary block on a slope is B.

Analysis of the Forces
There are three main forces acting on the block:

Weight (W or Fg): This force is due to gravity and always acts vertically downwards, from the center of mass of the object towards the center of the Earth. It does not act perpendicular to the slope, unless the slope is horizontal.

Normal Force : This is the force the surface exerts on the object to support it. It acts perpendicular to the surface of the slope. This force is a reaction to the component of the object's weight that pushes it into the slope.

Friction: Since the block is stationary on the slope, there is a static frictional force preventing it from sliding down. This force acts parallel to the surface of the slope and points upwards, opposing the component of gravity that is pulling the block downwards along the slope.

Why Diagram B is Correct
Diagram A is incorrect because the normal force is not perpendicular to the slope.

Diagram B correctly shows all three forces: the weight (W) acting vertically downwards, the normal force (N) acting perpendicular to the slope, and the frictional force (Ff) acting up the slope. The arrows are placed correctly.

Diagram C is incorrect there is no normal contact force.

Diagram D is incorrect because it shows the normal force acting vertically upwards, which is only true for a flat, horizontal surface.


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Question 5:

The correct answer is D. There is no resultant force.

Explanation

Since the block is stationary on the slope, it's in a state of equilibrium. This means all the forces acting on it are balanced. The resultant force, which is the vector sum of all the forces, must be zero.

The forces on the block are:

The downward component of the weight is balanced by the upward force of static friction. The perpendicular component of the weight is balanced by the normal force. Because all forces are balanced, the net force is zero, so there is no resultant force acting on the block.


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Question 6:

The correct statement is B. Normal force = m.g.cosθ.

Explanation
The normal force is the force exerted by a surface to prevent an object from penetrating it. On a horizontal surface, the normal force is equal to the object's weight (mg). However, on an inclined plane, the normal force is equal to the component of the weight that is perpendicular to the slope.

To find this component, we resolve the weight vector (mg), which acts vertically downwards, into two components:

The component of weight perpendicular to the slope is the force pushing the block into the slope. Using trigonometry, this component is equal to mg×cosθ.

Since the block is stationary and not moving perpendicular to the slope, the normal force (N) must be equal and opposite to this component of the weight.

Therefore, the normal force is given by the equation:
N=mgcosθ


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Question 7:

Calculation
To find the force of air resistance, we need to consider the net force acting on the stone. The net force is the sum of the downward force of gravity and the upward force of air resistance.

Calculate the force of gravity (Fg)

Fg=mass × acceleration due to gravity (g)
Assuming g≈9.8 m/s2
Fg=0.5 kg×9.8 m/s2 =4.9 N

Calculate the net force (Fnet):
The net force is what causes the stone to accelerate.
Fnet =mass×acceleration =0.5 kg×3 m/s
=1.5 N

Find the force of air resistance (Fair):
The forces are in opposite directions, so the net force is the difference between the force of gravity (downwards) and the force of air resistance (upwards).
Fnet =Fg - Fair

1.5 N=4.9 N−Fair


So Fair =4.9 N−1.5 N

Fair = 3.4 N

This value is approximately 3.5 N, which corresponds to answer B.


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Question 8:

To find the force, we use Hooke's Law, which states that the force applied to a spring is directly proportional to its extension. The formula for Hooke's Law is:
F=kx
Where:

F is the force applied (in Newtons, N)

k is the spring constant (in N/m)

x is the extension (in meters, m)

Calculation
First, we need to convert the extension from centimeters to meters.
5 cm=0.05 m

Now, we can plug the values into the formula:
F=(500 N/m)×(0.05 m)
F=25 N


This corresponds to answer D.

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Question 9:

Calculation

To find the extension, you need to use a variation of Hooke's Law (), where the force () on the spring is the weight of the added mass.

  1. Calculate the force (weight): We use for simplicity in these types of problems.

  2. Rearrange Hooke's Law to solve for extension ():

  3. Substitute the values and solve for :

  4. Convert the extension to centimeters:

    This corresponds to answer C.


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Question 10:

The correct answer is D. k = 250 N m⁻¹ and L = 27 cm.

Finding the Spring Constant (k)
Hooke's Law states that the force applied to a spring is proportional to its extension (F=kx). The force here is the weight of the mass (F=mg). The extension (x) is the difference between the stretched length and the natural length (x=stretched length−natural length).

Let L0 be the natural length of the spring.
We have two scenarios:

Scenario 1: A mass of 2 kg gives a length of 15 cm.
F1=m g=2 kg×10 N/kg =20 N
x1 =15 cm−L0

Using Hooke's Law: 20=k(15−L0) (Equation 1)

Scenario 2: A mass of 3 kg gives a length of 19 cm.
F2 =m g=3 kg×10 N/kg =30 N
x2=19 cm−L0

Using Hooke's Law: 30=k(19−L0) (Equation 2)

We now have a system of two equations with two unknowns (k and L0)
Let's divide Equation 2 by Equation 1 to eliminate k:
30/20 = [ k(19−L0) ] / [k(15−L0) ]

1.5= (19-L0) / (15 -L0)

1.5(15−L0)=19−L0

22.5−1.5L0=19−L0

3.5=0.5L0

L0 =7 cm

Now, substitute L0 back into Equation 1 to find k:
20=k(15−7)
20=k(8)
k= 20/8 ​
=2.5 N/cm

To convert the spring constant to N/m, we multiply by 100 (since there are 100 cm in a meter).
k=2.5 N/cm×100=250 N/m

Finding the Length for a 5 kg Mass
Now that we know k=250 N/m and the natural length L0 =7 cm, we can find the length when a 5 kg mass is added.

Calculate the force:
F =m g=5 kg×10 N/kg =50 N

Calculate the extension (x):
x= F/k
​ x= 50 / 250  ​
x =0.2 m

Convert the extension to cm:
x=0.2 m×100=20 cm

Find the final length (L):
L=L0+x
L=7 cm+20 cm=27 cm

The correct answer is D. k = 250 N m⁻¹ and L = 27 cm.


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