Physics Exam & Worksheet Study Guidance: Forces, Kinematics, and Dynamics

Q1. A person stands in an elevator moving upward at constant speed. What is the magnitude of the upward contact force, N, exerted by the elevator floor on the person (relative to the person's weight, W)?

Background

Topic: Newton's Laws of Motion (Forces in Equilibrium)

This question tests your understanding of the forces acting on a person in an elevator, specifically when the elevator moves at constant velocity. It focuses on the relationship between the normal force (contact force from the floor) and the gravitational force (weight).

Key Terms and Formulas:

• Normal Force ($N$): The upward force exerted by the floor.

• Weight ($W$): The downward force due to gravity, $W = mg$.

• Newton's First Law: If velocity is constant, the net force is zero.

Step-by-Step Guidance

1. Draw a free-body diagram for the person, showing all vertical forces (normal force up, weight down).

2. Recall that constant speed means zero acceleration, so the net force must be zero ($\sum F = 0$).

3. Set up the equation for forces: $N - W = 0$.

4. Rearrange to solve for $N$ in terms of $W$.

Try solving on your own before revealing the answer!

Final Answer: N = W

When the elevator moves at constant speed, the normal force equals the person's weight because there is no acceleration.

Q2. Now suppose the elevator is accelerating upward. How does the normal force compare to the weight of the person?

Background

Topic: Newton's Second Law (Forces with Acceleration)

This question explores how the normal force changes when the elevator accelerates upward, requiring you to apply Newton's second law to a non-equilibrium situation.

Key Terms and Formulas:

• Newton's Second Law: $\sum F = ma$

• Normal Force ($N$), Weight ($W$), Acceleration ($a$)

Step-by-Step Guidance

1. Draw a free-body diagram for the person, showing $N$ upward and $W$ downward.

2. Write Newton's second law for the vertical direction: $N - W = ma$ (where $a$ is upward acceleration).

3. Rearrange to solve for $N$ in terms of $W$ and $a$.

4. Think about whether $N$ is greater than, less than, or equal to $W$ when $a > 0$.

Try solving on your own before revealing the answer!

Final Answer: N > W

When the elevator accelerates upward, the normal force must be greater than the weight to provide the upward acceleration.

Q3. In a tilted coordinate system, the acceleration vector is always in the x-direction. The coordinate axes are at an angle θ to the horizontal. What are the x- and y-components of the acceleration?

Background

Topic: Kinematics in Rotated Coordinate Systems

This question tests your ability to resolve vectors (specifically acceleration) into components along tilted axes, which is common in inclined plane problems.

Key Terms and Formulas:

• Component of acceleration: $a_x = a \cos \theta$, $a_y = a \sin \theta$ (depending on axis orientation)

• Trigonometric relationships for vector decomposition

Step-by-Step Guidance

1. Draw the coordinate axes, labeling the angle θ from the horizontal.

2. Express the acceleration vector along the x-axis of the tilted system.

3. Use trigonometry to write the x- and y-components in terms of $a$ and $\theta$.

4. Check the direction (sign) of each component based on the diagram.

Try solving on your own before revealing the answer!

Final Answer: $a_x = a \cos \theta$, $a_y = a \sin \theta$

The components depend on the orientation of the axes and the direction of the acceleration vector.

Q4. A mass is on a frictionless inclined plane. If released from rest, in which direction is the acceleration?

Background

Topic: Dynamics on Inclined Planes

This question asks you to identify the direction of acceleration for an object on a frictionless incline, a classic application of Newton's laws and vector decomposition.

Key Terms and Formulas:

• Gravity acts vertically downward: $g$

• On a frictionless incline, only the component of gravity parallel to the surface causes acceleration.

Step-by-Step Guidance

1. Draw a free-body diagram for the mass, showing gravity and the normal force.

2. Decompose the gravitational force into components parallel and perpendicular to the incline.

3. Recognize that the normal force cancels the perpendicular component, leaving only the parallel component to cause acceleration.

4. State the direction of the net force and thus the acceleration (down the incline).

Try solving on your own before revealing the answer!

Final Answer: Down the incline

The acceleration is directed parallel to the surface, down the incline, due to the component of gravity along the plane.

Q5. A mass is pulled along a rough table at constant velocity with an external force F at some angle above the horizontal. The magnitudes of the forces on the free-body diagram have been drawn correctly, but the directions of the forces are correct. Which statement below must be true?

Background

Topic: Forces and Equilibrium with Friction

This question tests your understanding of equilibrium conditions when an object is pulled at an angle with friction present.

Key Terms and Formulas:

• Constant velocity: Net force is zero in all directions.

• Forces: Applied force $F$, friction $f$, normal force $N$, weight $mg$.

• Equilibrium equations: $\sum F_x = 0$, $\sum F_y = 0$

Step-by-Step Guidance

1. Write the force balance equations for the x- and y-directions.

2. In the x-direction: $F \cos \theta - f = 0$ (applied force balances friction).

3. In the y-direction: $N + F \sin \theta - mg = 0$ (vertical forces balance).

4. Identify which relationships must be true for equilibrium.

Try solving on your own before revealing the answer!

Final Answer: $F \cos \theta = f$ and $N + F \sin \theta = mg$

These equations ensure the net force is zero in both directions, satisfying equilibrium for constant velocity.