Newtons Second Law exercise - Free Printable

Problem Analysis:

The worksheet provided is about Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this is expressed as:

\[
F = m \cdot a
\]

where:
- \( F \) is the net force (in Newtons, N),
- \( m \) is the mass (in kilograms, kg),
- \( a \) is the acceleration (in meters per second squared, \( \text{m/s}^2 \)).

We will solve each part of the worksheet step by step.

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Part A: Select the correct term to complete these sentences

1. When a resultant force acts on a mass, its ______ increases in the direction of that force. This change of speed is called ______. The greater the force acting on the mass, the ______ its acceleration.

- First blank: When a force acts on a mass, its velocity increases.
- Second blank: The change in velocity is called acceleration.
- Third blank: The greater the force, the greater the acceleration.

Completed sentence:
"When a resultant force acts on a mass, its velocity increases in the direction of that force. This change of speed is called acceleration. The greater the force acting on the mass, the greater its acceleration."

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Part B: Answer these questions to 2 decimal places and select the appropriate units

#### 1. What force is needed to accelerate a 23Kg mass at a rate of 12 m/s²?

Using Newton's Second Law:
\[
F = m \cdot a
\]
Substitute the given values:
\[
m = 23 \, \text{kg}, \quad a = 12 \, \text{m/s}^2
\]
\[
F = 23 \cdot 12 = 276 \, \text{N}
\]

Answer: \( 276.00 \, \text{N} \)

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#### 2. A 50g toy car is pushed with a force of 1N. Calculate its acceleration.

First, convert the mass from grams to kilograms:
\[
50 \, \text{g} = 0.05 \, \text{kg}
\]
Using Newton's Second Law:
\[
a = \frac{F}{m}
\]
Substitute the given values:
\[
F = 1 \, \text{N}, \quad m = 0.05 \, \text{kg}
\]
\[
a = \frac{1}{0.05} = 20 \, \text{m/s}^2
\]

Answer: \( 20.00 \, \text{m/s}^2 \)

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#### 3. What is the weight (a force) of a 1Kg mass? (The acceleration due to gravity is 10 m/s²)

Weight is the force exerted on an object due to gravity. It can be calculated using:
\[
W = m \cdot g
\]
where \( g \) is the acceleration due to gravity. Substitute the given values:
\[
m = 1 \, \text{kg}, \quad g = 10 \, \text{m/s}^2
\]
\[
W = 1 \cdot 10 = 10 \, \text{N}
\]

Answer: \( 10.00 \, \text{N} \)

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#### 4. What mass has a motorbike, if a force of 1,800N from the engine gives it an acceleration of 6 m/s²?

Using Newton's Second Law:
\[
m = \frac{F}{a}
\]
Substitute the given values:
\[
F = 1,800 \, \text{N}, \quad a = 6 \, \text{m/s}^2
\]
\[
m = \frac{1,800}{6} = 300 \, \text{kg}
\]

Answer: \( 300.00 \, \text{kg} \)

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#### 5. A firework rocket has a mass of 120g. Immediately after lift-off, the force created by the burning powder is 3.9N, but the friction due to air resistance equals 0.66N:

##### a. Calculate the acceleration of the rocket at this time.

The net force acting on the rocket is the difference between the thrust force and the air resistance:
\[
F_{\text{net}} = F_{\text{thrust}} - F_{\text{air resistance}}
\]
Substitute the given values:
\[
F_{\text{thrust}} = 3.9 \, \text{N}, \quad F_{\text{air resistance}} = 0.66 \, \text{N}
\]
\[
F_{\text{net}} = 3.9 - 0.66 = 3.24 \, \text{N}
\]

Convert the mass from grams to kilograms:
\[
120 \, \text{g} = 0.12 \, \text{kg}
\]

Using Newton's Second Law:
\[
a = \frac{F_{\text{net}}}{m}
\]
Substitute the values:
\[
F_{\text{net}} = 3.24 \, \text{N}, \quad m = 0.12 \, \text{kg}
\]
\[
a = \frac{3.24}{0.12} = 27 \, \text{m/s}^2
\]

Answer: \( 27.00 \, \text{m/s}^2 \)

##### b. As the speed of the rocket increases, the size of the friction force due to air resistance also increases. How would this affect the acceleration of the rocket?

As the speed of the rocket increases, the air resistance force (\( F_{\text{air resistance}} \)) also increases. This reduces the net force acting on the rocket:
\[
F_{\text{net}} = F_{\text{thrust}} - F_{\text{air resistance}}
\]
Since \( F_{\text{air resistance}} \) is increasing, \( F_{\text{net}} \) decreases. According to Newton's Second Law (\( a = \frac{F_{\text{net}}}{m} \)), a decrease in \( F_{\text{net}} \) results in a decrease in acceleration. Therefore, the acceleration of the rocket will decrease as the speed increases.

##### c. The acceleration was not found to change as much as anticipated. What factor could be causing this to occur?

If the acceleration does not decrease as much as expected despite the increase in air resistance, there could be several factors:
1. Thrust Increase: The thrust force (\( F_{\text{thrust}} \)) might be increasing as the rocket burns more fuel, compensating for the increased air resistance.
2. Drag Coefficient: The drag coefficient of the rocket might be lower than expected, meaning the air resistance does not increase as rapidly with speed.
3. Measurement Error: There could be errors in measuring the forces or accelerations.
4. Rocket Design: The shape or design of the rocket might be optimized to reduce air resistance effectively.

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Final Answers:

1. \( 276.00 \, \text{N} \)
2. \( 20.00 \, \text{m/s}^2 \)
3. \( 10.00 \, \text{N} \)
4. \( 300.00 \, \text{kg} \)
5. a. \( 27.00 \, \text{m/s}^2 \)
b. The acceleration decreases as air resistance increases.
c. Possible factors include increasing thrust, lower drag coefficient, measurement errors, or rocket design optimization.

Boxed Final Answer:
\[
\boxed{276.00 \, \text{N}, 20.00 \, \text{m/s}^2, 10.00 \, \text{N}, 300.00 \, \text{kg}, 27.00 \, \text{m/s}^2}
\]