Physics with Calculus: Step-by-Step Guidance for Test 2 Review

Q1. Convert 300 lbs. into Newtons. (Note: 1 pound = 4.45 N)

Background

Topic: Unit Conversions (Forces)

This question tests your ability to convert between different units of force, specifically from pounds (imperial) to newtons (SI).

Key Terms and Formulas:

• 1 pound (lb) = 4.45 newtons (N)

• Force in newtons: $F_{N} = F_{lb} \times \text{conversion factor}$

Step-by-Step Guidance

1. Write down the given value: $F_{lb} = 300$ lbs.

2. Identify the conversion factor: $1$lb$= 4.45$ N.

3. Set up the conversion: $F_{N} = 300 \times 4.45$

4. Multiply to find the force in newtons, but do not calculate the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 1335 N

$F_{N} = 300 \times 4.45 = 1335$ N

So, 300 lbs is equivalent to 1335 newtons.

Q2. Convert 9.80 m/s2 into ft/s2. (Note: 1 m = 39.37 in)

Background

Topic: Unit Conversions (Acceleration)

This question tests your ability to convert acceleration from SI units (meters per second squared) to imperial units (feet per second squared), requiring multiple steps.

Key Terms and Formulas:

• 1 m = 39.37 in

• 1 ft = 12 in

• To convert m/s2 to ft/s2:

  $a_{ft/s^2} = a_{m/s^2} \times \frac{39.37}{12}$

Step-by-Step Guidance

1. Start with the given acceleration: $a_{m/s^2} = 9.80$ m/s2.

2. Convert meters to inches: $9.80$ m/s2 $\times 39.37$ in/m.

3. Convert inches to feet: Divide by 12 in/ft.

4. Set up the full conversion: $a_{ft/s^2} = 9.80 \times \frac{39.37}{12}$

5. Multiply and simplify, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 32.2 ft/s2

$a_{ft/s^2} = 9.80 \times \frac{39.37}{12} \approx 32.2$ ft/s2

This is the standard value for acceleration due to gravity in feet per second squared.

Q3. Algebraic manipulation: Given $\Sigma F_y = n - mg \cos \theta = ma_y$, solve for $a_y$ and $\theta$.

Background

Topic: Algebraic Manipulation of Newton's Second Law

This question tests your ability to rearrange equations to solve for a specific variable, a key skill in physics problem-solving.

Key Terms and Formulas:

• $\Sigma F_y$ = sum of forces in the y-direction

• $n$ = normal force

• $m$ = mass

• $g$ = acceleration due to gravity

• $\theta$ = angle

• $a_y$ = acceleration in the y-direction

Step-by-Step Guidance

1. Start with the equation: $n - mg \cos \theta = ma_y$

2. To solve for $a_y$, isolate $a_y$ on one side by dividing both sides by $m$.

3. To solve for $\theta$, rearrange the equation to isolate $\cos \theta$ and then $\theta$.

4. Write the expressions for $a_y$ and $\theta$ in terms of the other variables, but do not substitute values or simplify fully yet.

Try solving on your own before revealing the answer!

Final Answer:

$a_y = \frac{n}{m} - g \cos \theta$

$\theta = \cos^{-1}\left(\frac{n - m a_y}{m g}\right)$

These are the rearranged forms for $a_y$ and $\theta$.

Q4. Vector Components: The x-component of the tension in a cable is -1200 N and the y-component is 600 N.

Background

Topic: Vector Addition and Components

This question tests your understanding of how to represent vectors graphically and analytically, and how to find the magnitude and direction from components.

Key Terms and Formulas:

• Vector components: $T_x$, $T_y$

• Magnitude: $|\vec{T}| = \sqrt{T_x^2 + T_y^2}$

• Direction: $\theta = \tan^{-1}\left(\frac{T_y}{T_x}\right)$

Step-by-Step Guidance

1. Sketch the vector components: Draw $T_x$ to the left (negative x-direction) and $T_y$ upward (positive y-direction).

2. Draw the resultant vector from the origin to the point ($T_x$, $T_y$).

3. Calculate the magnitude using the Pythagorean theorem: $|\vec{T}| = \sqrt{(-1200)^2 + (600)^2}$

4. Find the angle with respect to the x-axis: $\theta = \tan^{-1}\left(\frac{600}{-1200}\right)$

5. Do not compute the final values yet; set up the expressions for the student to finish.

Try solving on your own before revealing the answer!

Final Answer:

Magnitude: $|\vec{T}| = 1342$ N

Angle: $\theta = 153.4^\circ$ (measured from the positive x-axis, counterclockwise)

The sum of these components is called the resultant vector.

Q5. Analytically add the following vectors: <410 N, 111°> + <683 N, 280°> = < _______, _______ >

Background

Topic: Analytical Vector Addition

This question tests your ability to add vectors given in magnitude and direction (polar form) by converting to components, adding, and converting back.

Key Terms and Formulas:

• Component form: $A_x = A \cos \theta$, $A_y = A \sin \theta$

• Resultant: $R_x = A_x + B_x$, $R_y = A_y + B_y$

• Magnitude: $R = \sqrt{R_x^2 + R_y^2}$

• Direction: $\phi = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Step-by-Step Guidance

1. Convert each vector to x and y components using $A_x = A \cos \theta$, $A_y = A \sin \theta$.

2. Add the x-components and y-components to get $R_x$ and $R_y$.

3. Calculate the magnitude of the resultant vector: $R = \sqrt{R_x^2 + R_y^2}$

4. Find the direction: $\phi = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

5. Set up the expressions, but do not compute the final values yet.

Try solving on your own before revealing the answer!

Final Answer:

Resultant: $\langle 273.5 \text{ N}, 123.2^\circ \rangle$

After converting to components, adding, and converting back, this is the resultant vector in magnitude and direction.

Q6. List examples for each of Newton’s 3 laws you have experienced today.

Background

Topic: Newton's Laws of Motion

This question tests your conceptual understanding of Newton's 1st, 2nd, and 3rd laws by asking for real-life examples.

Key Terms:

• Newton's 1st Law: Law of Inertia

• Newton's 2nd Law: $F = ma$

• Newton's 3rd Law: Action-Reaction

Step-by-Step Guidance

1. Recall the statement of each law.

2. Think of a daily activity that illustrates each law (e.g., riding in a car, pushing a door, jumping off a step).

3. Write a brief description for each example, connecting it to the law's principle.

Try coming up with your own examples before revealing the answer!

Final Answer:

• 1st Law: When a car suddenly stops, your body lurches forward due to inertia.

• 2nd Law: Pushing a shopping cart: the harder you push (more force), the faster it accelerates (more $a$).

• 3rd Law: Jumping off a boat: you push the boat backward as you move forward.

Q7. Without altering their eating habits or exercising, what can a person do to increase or decrease their weight? Mass?

Background

Topic: Mass vs. Weight

This question tests your understanding of the difference between mass (amount of matter) and weight (force due to gravity).

Key Terms:

• Mass ($m$): Amount of matter, measured in kg.

• Weight ($W$): Force due to gravity, $W = mg$.

Step-by-Step Guidance

1. Recall that mass is constant unless you add or remove matter.

2. Weight depends on the local gravitational acceleration ($g$).

3. Think of ways to change $g$ (e.g., go to a different planet or altitude).

4. Consider if mass can change without eating/exercising.

Try answering before revealing the answer!

Final Answer:

To change weight without changing mass, go to a place with different gravity (e.g., the Moon or another planet). Mass remains unchanged unless you add or remove matter from your body.

Q8. On Mars (mass $6.42 \times 10^{23}$ kg, radius $3.37 \times 10^6$ m), your space suit has a mass of 136.4 kg.

Background

Topic: Newton's Law of Universal Gravitation and Weight

This question tests your ability to calculate gravitational acceleration, weight on different planets, and orbital speed.

Key Terms and Formulas:

• Gravitational acceleration: $g = \frac{GM}{r^2}$

• Weight: $W = mg$

• Orbital speed: $v = \sqrt{\frac{GM}{r}}$

• $G = 6.674 \times 10^{-11}$ N m2/kg2

Step-by-Step Guidance

1. For Mars' gravity, use $g = \frac{GM}{r^2}$ with Mars' mass and radius.

2. For the weight of the suit on Earth, use $W = mg$ with $g = 9.80$ m/s2.

3. For the weight on Mars, use the $g$ you calculated for Mars.

4. For orbital speed, add 50 km to Mars' radius and use $v = \sqrt{\frac{GM}{r}}$.

5. Set up each calculation, but do not compute the final values yet.

Try solving on your own before revealing the answer!

Final Answer:

• A) $g_{Mars} = 3.76$ m/s2

• B) $W_{Earth} = 1338$ N

• C) $W_{Mars} = 513$ N

• D) $v_{orbit} = 3370$ m/s

Each value is found by plugging into the respective formula with the given data.

Q9. An ant (mass 3.0 mg, length 5.0 mm, width 1.0 mm) is dropped from 100 m. A) What is its final speed upon impact if there is no air resistance?

Background

Topic: Kinematics (Free Fall)

This question tests your ability to use kinematic equations to find the final speed of an object in free fall from a given height, neglecting air resistance.

Key Terms and Formulas:

• Initial velocity ($v_0$): 0 (dropped)

• Displacement ($h$): 100 m

• Acceleration ($g$): 9.80 m/s2

• Final speed: $v = \sqrt{2gh}$

Step-by-Step Guidance

1. Identify the known values: $h = 100$ m, $g = 9.80$ m/s2, $v_0 = 0$.

2. Use the kinematic equation for final velocity: $v^2 = v_0^2 + 2gh$.

3. Since $v_0 = 0$, simplify to $v = \sqrt{2gh}$.

4. Set up the calculation: $v = \sqrt{2 \times 9.80 \times 100}$, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

$v = 44.3$ m/s

This is the speed the ant would reach if dropped from 100 m with no air resistance.

Q10. What is the net force and acceleration of a 30 kg mass with forces 300 N, 200 N, and 150 N acting on it?

Background

Topic: Newton's Second Law (Net Force and Acceleration)

This question tests your ability to find the net force acting on an object and use it to calculate acceleration.

Key Terms and Formulas:

• Net force: $F_{net} = \sum F$

• Newton's Second Law: $F_{net} = ma$

• Acceleration: $a = \frac{F_{net}}{m}$

Step-by-Step Guidance

1. Add up all the forces, taking direction into account (are they all in the same direction or opposite?).

2. Calculate the net force: $F_{net} = 300 + 200 + 150$ (if all in the same direction).

3. Use $a = \frac{F_{net}}{m}$ with $m = 30$ kg.

4. Set up the calculation for acceleration, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

Net force: $F_{net} = 650$ N

Acceleration: $a = 21.7$ m/s2

Both values are found by summing the forces and dividing by the mass.

Q11. A 95 kg man stands on a scale in an elevator. What is the scale reading when the elevator moves upward at a constant speed of 6 m/s?

Background

Topic: Apparent Weight in Non-Inertial Frames

This question tests your understanding of normal force (apparent weight) in an accelerating or moving elevator.

Key Terms and Formulas:

• Apparent weight: $N$ (normal force)

• At constant speed: $a = 0$

• Normal force: $N = mg$

Step-by-Step Guidance

1. Recognize that constant speed means no acceleration ($a = 0$).

2. The only forces are gravity and the normal force.

3. Set up the equation: $N = mg$.

4. Plug in $m = 95$ kg and $g = 9.80$ m/s2, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

Scale reading: $N = 931$ N

At constant speed, the scale reads the true weight.

Q12. Calculate the maximum deceleration of a car heading down a 20° slope on dry concrete (static friction applies, μ given).

Background

Topic: Friction and Inclined Planes

This question tests your ability to analyze forces on an inclined plane and use friction to find maximum deceleration.

Key Terms and Formulas:

• Static friction: $f_{max} = \mu_s N$

• Normal force: $N = mg \cos \theta$

• Down-slope component: $mg \sin \theta$

• Maximum deceleration: $a_{max} = g(\mu_s \cos \theta - \sin \theta)$

Step-by-Step Guidance

1. Draw a free-body diagram showing gravity, normal force, and friction.

2. Write the equation for forces along the slope: $f_{max} - mg \sin \theta = ma$.

3. Substitute $f_{max} = \mu_s mg \cos \theta$.

4. Solve for $a_{max}$: $a_{max} = g(\mu_s \cos \theta - \sin \theta)$.

5. Set up the calculation with the given values, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

Maximum deceleration: $a_{max} = 2.0$ m/s2

This is found by plugging the values for $g$, $\mu_s$, and $\theta$ into the formula.

Q13. Find the acceleration of a 9 kg object moving to the right with a 12 N force, $\mu = 0.41$, $g = -9.80$ m/s2, $\theta = 41^\circ$.

Background

Topic: Newton's Second Law with Friction and Inclined Planes

This question tests your ability to analyze forces on an inclined plane with friction and calculate acceleration.

Key Terms and Formulas:

• Friction force: $f = \mu N$

• Normal force: $N = mg \cos \theta$

• Down-slope component: $mg \sin \theta$

• Net force: $F_{net} = F_{applied} - f - mg \sin \theta$

• Acceleration: $a = \frac{F_{net}}{m}$

Step-by-Step Guidance

1. Calculate the normal force: $N = mg \cos \theta$.

2. Find the friction force: $f = \mu N$.

3. Calculate the component of gravity down the slope: $mg \sin \theta$.

4. Set up the net force equation: $F_{net} = 12 - f - mg \sin \theta$.

5. Set up the acceleration formula: $a = \frac{F_{net}}{9}$, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

Acceleration: $a = -5.1$ m/s2

The negative sign indicates the object is decelerating or the net force is opposite the direction of motion.

Q14. Determine the acceleration of mass $m_1$ (51 kg) moving upward, with $m_2 = 101$ kg on a 72° incline, $\mu = 0.14$.

Background

Topic: Atwood Machine with Friction and Inclined Plane

This question tests your ability to analyze a two-mass system with pulleys, friction, and an incline, and to solve for acceleration.

Key Terms and Formulas:

• Forces on $m_1$ (vertical): $T - m_1g = m_1a$

• Forces on $m_2$ (incline): $m_2g \sin \theta - T - f = m_2a$

• Friction: $f = \mu m_2g \cos \theta$

• Combine equations to solve for $a$.

Step-by-Step Guidance

1. Write the force equations for both masses, including friction for $m_2$.

2. Express the tension $T$ in terms of $a$ for both equations.

3. Set the two expressions for $T$ equal to each other and solve for $a$.

4. Plug in the given values for $m_1$, $m_2$, $\theta$, $\mu$, and $g$.

5. Set up the final equation for $a$, but do not compute the value yet.

Try solving on your own before revealing the answer!

Final Answer:

Acceleration: $a = 2.1$ m/s2

This is found by solving the system of equations with the given values.