Physics Exam Study Guide: Step-by-Step Guidance with Relevant Images

Q2. Newton’s 2nd Law with Friction: In the picture below, a horizontal force of magnitude 30.0 N is pushing two blocks to the right. Block A has mass $m_A = 3.00$ kg and block B has mass $m_B = 2.00$ kg. The coefficient of kinetic friction between the blocks and the floor is $\mu_k = 0.2$. What is the magnitude of the force that block A exerts on block B?

Background

Topic: Newton's Laws of Motion & Friction

This question tests your understanding of Newton's second law, free-body diagrams, and the effects of friction when multiple objects are pushed together.

Key Terms and Formulas:

• Newton's Second Law: $F_{\text{net}} = m a$

• Kinetic friction force: $f_k = \mu_k F_N$

• Free-body diagram: Analyze forces acting on each block.

Step-by-Step Guidance

1. Draw a free-body diagram for both blocks. Identify all forces: applied force $F$, friction force, and the contact force between blocks A and B.

2. Calculate the total friction force acting on both blocks: $f_k = \mu_k (m_A + m_B) g$.

3. Apply Newton's second law to the system (both blocks together) to find the acceleration: $F - f_k = (m_A + m_B) a$.

4. Now, focus on block B. The only horizontal force acting on block B is the force from block A (call it $F_{AB}$), minus friction. Set up Newton's second law for block B: $F_{AB} - f_{k,B} = m_B a$.

Try solving on your own before revealing the answer!

Final Answer: 12.0 N

$\text{Using the steps above, the force that block A exerts on block B is:}$

$F_{AB} = m_B a + f_{k,B}$

$a$ is found from the total system, and $f_{k,B} = \mu_k m_B g$. Plug in the values to get $F_{AB} = 12.0$ N.

This is the contact force between the two blocks, accounting for friction and acceleration.

Q6. Rotational Dynamics about a Fixed Axis: A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?

Background

Topic: Rotational Dynamics & Moment of Inertia

This question tests your ability to relate linear and rotational motion, and to use Newton's second law for rotation to solve for the moment of inertia.

Key Terms and Formulas:

• Moment of inertia ($I$): Resistance to rotational acceleration.

• Newton's second law for rotation: $\tau = I \alpha$

• Relationship between linear and angular acceleration: $a = r \alpha$

• Tension in the rope provides torque: $\tau = T r$

Step-by-Step Guidance

1. Calculate the linear acceleration of the block using kinematics: $s = \frac{1}{2} a t^2$.

2. Relate the linear acceleration to angular acceleration: $a = r \alpha$.

3. Set up the torque equation: $\tau = T r = I \alpha$.

4. Combine the equations to solve for $I$: $I = \frac{T r^2}{a}$.

Try solving on your own before revealing the answer!

Final Answer: 4.3 kg·m2

$\text{Using the calculated acceleration and the torque equation, the moment of inertia is:}$

$I = \frac{T r^2}{a}$

Plug in $T = 20$ N, $r = 0.40$ m, and $a$ from kinematics to get $I = 4.3$ kg·m2.

Q7. Static Equilibrium: A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall. The ladder forms an angle of 53° above horizontal. The coefficient of static friction between the floor and the ladder is 0.28. What distance, measured along the ladder from the bottom, can a 60-kg person climb before the ladder starts to slip?

Background

Topic: Static Equilibrium & Friction

This question tests your ability to analyze forces and torques in a system at rest, and to determine the maximum safe position before slipping occurs.

Key Terms and Formulas:

• Static equilibrium: $\sum F = 0$, $\sum \tau = 0$

• Maximum static friction: $f_{s,\text{max}} = \mu_s F_N$

• Torque: $\tau = r F \sin \theta$

Step-by-Step Guidance

1. Draw a free-body diagram showing all forces: weight of ladder, weight of person, normal force, friction force, and reaction force from the wall.

2. Write the equilibrium equations for forces in horizontal and vertical directions.

3. Set up the torque equation about the base of the ladder to relate the position of the person to the maximum static friction.

4. Solve for the distance along the ladder where the torque due to the person’s weight causes the friction force to reach its maximum value.

Try solving on your own before revealing the answer!

Final Answer: 1.73 m

$\text{Using the torque and force balance equations, the maximum distance is:}$

$d = 1.73$ m

This is the farthest the person can climb before the ladder slips, based on static friction and equilibrium.

Q9. Pressure in a Fluid: The two water reservoirs shown in the figure are open to the atmosphere, and the water has density 1000 kg/m3. The manometer contains incompressible mercury with a density of 13,600 kg/m3. What is the difference in elevation h if the manometer reading m is 25.0 cm?

Background

Topic: Fluid Statics & Pressure Measurement

This question tests your understanding of pressure differences in fluids, manometers, and the use of density in calculating pressure.

Key Terms and Formulas:

• Pressure difference: $\Delta P = \rho g h$

• Manometer principle: relates height of fluid columns to pressure difference

• Density ($\rho$): mass per unit volume

Step-by-Step Guidance

1. Write the pressure difference equation for the manometer: $\Delta P = \rho_{\text{Hg}} g m$.

2. Set up the equation relating the pressure difference to the height difference $h$ in the water reservoirs: $\Delta P = \rho_{\text{water}} g h$.

3. Equate the two expressions for $\Delta P$ and solve for $h$.

4. Plug in the values for densities, $g$, and $m$ to set up the calculation for $h$.

Try solving on your own before revealing the answer!

Final Answer: 3.15 m

$\text{Using the pressure difference equations, the elevation difference is:}$

$h = \frac{\rho_{\text{Hg}} m}{\rho_{\text{water}}}$

Plug in the values to get $h = 3.15$ m.

Q11. Bernoulli's Principle: A paint sprayer pumps air through a constriction in a 2.50-cm diameter pipe, as shown in the figure. The flow causes the pressure in the constricted area to drop and paint rises up the feed tube and enters the air stream. The speed of the air stream in the 2.50-cm diameter sections is 5.00 m/s. The density of the air is 1.29 kg/m3, and the density of the paint is 1200 kg/m3. We can treat the air and paint as incompressible ideal fluids. What is the maximum diameter of the constriction that will allow the sprayer to operate?

Background

Topic: Fluid Dynamics & Bernoulli's Principle

This question tests your ability to apply Bernoulli's equation and continuity equation to determine the conditions for fluid flow and pressure differences.

Key Terms and Formulas:

• Bernoulli's equation: $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$

• Continuity equation: $A_1 v_1 = A_2 v_2$

• Pressure difference needed to lift paint: $\Delta P = \rho_{\text{paint}} g h$

Step-by-Step Guidance

1. Use the continuity equation to relate the velocities in the wide and narrow sections: $A_1 v_1 = A_2 v_2$.

2. Apply Bernoulli's equation between the wide and narrow sections to find the pressure drop.

3. Set the pressure drop equal to the pressure needed to lift the paint: $\Delta P = \rho_{\text{paint}} g h$.

4. Solve for the maximum diameter of the constriction using the equations above.

Try solving on your own before revealing the answer!

Final Answer: 8.07 mm

$\text{Using Bernoulli's and continuity equations, the maximum diameter is:}$

$d = 8.07$ mm

This ensures the pressure drop is sufficient to lift the paint into the air stream.

Q13. Mathematics of Traveling Waves: The figure shows the displacement y of a traveling wave at a given position as a function of time and the displacement of the same wave at a given time as a function of position. How fast is the wave traveling?

Background

Topic: Wave Motion & Speed Calculation

This question tests your ability to interpret wave graphs and use the relationship between wavelength, period, and speed.

Key Terms and Formulas:

• Wave speed: $v = \frac{\lambda}{T}$

• Wavelength ($\lambda$): distance between two consecutive points in phase

• Period ($T$): time for one complete cycle

Step-by-Step Guidance

1. From the position graph, determine the wavelength $\lambda$ by measuring the distance between two peaks.

2. From the time graph, determine the period $T$ by measuring the time between two peaks.

3. Use the formula $v = \frac{\lambda}{T}$ to set up the calculation for wave speed.

Try solving on your own before revealing the answer!

Final Answer: 0.75 m/s

$\text{Using the measured wavelength and period, the wave speed is:}$

$v = \frac{\lambda}{T} = 0.75$ m/s

This is the speed at which the wave travels along the string.

Q24. Saturated Vapor Pressure and Humidity: The saturated vapor pressure for water is given in the table to the right. On a cold winter day, the outside temperature is 0°C with a relative humidity of 60%. The heater of your apartment works by drawing cold air from the outside and heating it up to a comfortable temperature of 20°C. As a result, the air becomes very dry. How much water do you need to let evaporate if you’d like to raise the humidity of your bedroom (with a volume of 30 m3) to a comfortable 40%? (The molar mass of water is 18 g/mol)

Background

Topic: Vapor Pressure, Humidity, and Phase Changes

This question tests your understanding of relative humidity, vapor pressure, and the calculation of mass of water needed to achieve a desired humidity level.

Key Terms and Formulas:

• Relative humidity: $\text{RH} = \frac{P_{\text{actual}}}{P_{\text{sat}}} \times 100\%$

• Mass of water vapor: $m = n \times M$

• Ideal gas law: $P V = n R T$

Step-by-Step Guidance

1. Find the saturated vapor pressure at 20°C from the table.

2. Calculate the partial pressure of water vapor needed for 40% relative humidity at 20°C.

3. Use the ideal gas law to find the number of moles of water vapor needed in the room.

4. Convert moles to mass using the molar mass of water.

Try solving on your own before revealing the answer!

Final Answer: 125 g

$\text{Using the vapor pressure table and ideal gas law, the mass of water needed is:}$

$m = 125$ g

This is the amount of water you need to evaporate to reach 40% humidity in your room.