Physics 2A-32211 Exam #2 Review Exercises – Step-by-Step Study Guidance

Q1. A 60.5 kg student standing on a scale in an elevator notices that the scale reads 621 N. From this information, the student knows that the elevator must move:

Background

Topic: Newton's Laws of Motion (Forces in Non-Inertial Reference Frames)

This question tests your understanding of apparent weight and how it changes when an elevator accelerates upward or downward.

Key Terms and Formulas:

• Apparent weight: The normal force exerted by the scale, which can differ from true weight if the elevator accelerates.

• True weight: $W = mg$

• Normal force in an accelerating elevator: $N = m(g + a)$ (upward acceleration), $N = m(g - a)$ (downward acceleration)

Step-by-Step Guidance

1. Calculate the student's true weight using $W = mg$, where $m = 60.5$ kg and $g = 9.8$ m/s2.

2. Compare the scale reading (apparent weight) to the true weight. If the scale reads more than the true weight, the elevator is accelerating upward; if less, downward.

3. Set up the equation for the normal force: $N = m(g + a)$, and solve for the acceleration $a$ using the scale reading.

4. Interpret the sign of $a$ to determine the direction of acceleration.

Try solving on your own before revealing the answer!

Q2. A 1.0-kg slide down a frictionless plane inclined at an angle of 30 degrees with respect to the horizontal. Find the acceleration of the box.

Background

Topic: Newton's Second Law, Inclined Planes

This question tests your ability to resolve forces on an inclined plane and calculate acceleration.

Key Terms and Formulas:

• Gravity force: $F_g = mg$

• Component of gravity along the incline: $F_{\text{parallel}} = mg \sin \theta$

• Newton's Second Law: $F = ma$

Step-by-Step Guidance

1. Identify the forces acting on the box: gravity and normal force.

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

3. Write the equation for the net force along the incline: $F_{\text{net}} = mg \sin \theta$.

4. Apply Newton's Second Law: $a = \frac{F_{\text{net}}}{m}$.

Try solving on your own before revealing the answer!

Q3. Two blocks, with masses 60.0 kg and 40.0 kg, are connected by a rope and pulled by a force of 200 N as shown in the figure. The coefficient of kinetic friction between the blocks and the surface is 0.10. Find the acceleration of the blocks and the tension in the rope.

Background

Topic: Newton's Second Law, Friction, Systems of Connected Objects

This question tests your ability to analyze forces in a system of connected masses with friction.

Key Terms and Formulas:

• Kinetic friction: $f_k = \mu_k N$

• Newton's Second Law: $F = ma$

• System mass: $m_{\text{total}} = m_1 + m_2$

Step-by-Step Guidance

1. Calculate the total mass: $m_{\text{total}} = 60.0 + 40.0$ kg.

2. Find the normal force for each block and calculate the total friction force: $f_k = \mu_k N_{\text{total}}$.

3. Set up the net force equation: $F_{\text{net}} = F_{\text{pull}} - f_k$.

4. Apply Newton's Second Law to find the acceleration: $a = \frac{F_{\text{net}}}{m_{\text{total}}}$.

Try solving on your own before revealing the answer!

Q4. A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant, at which position is the tension in the cord connected to the ball the greatest?

Background

Topic: Circular Motion, Centripetal Force

This question tests your understanding of forces in vertical circular motion and how tension varies with position.

Key Terms and Formulas:

• Centripetal force: $F_c = \frac{mv^2}{r}$

• Tension at top and bottom: $T_{\text{top}} = F_c - mg$, $T_{\text{bottom}} = F_c + mg$

Step-by-Step Guidance

1. Identify the forces acting on the ball at the top and bottom of the circle.

2. Write the expressions for tension at both positions using the formulas above.

3. Compare the values to determine where tension is greatest.

Try solving on your own before revealing the answer!

Q5. Suppose you lift a suitcase from the floor to a table. The work you do on the suitcase depends only on the difference in height between the initial and final positions, not on the path taken. Explain why.

Background

Topic: Work and Energy, Conservative Forces

This question tests your understanding of conservative forces and path independence of work done by gravity.

Key Terms and Formulas:

• Work done by gravity: $W = mgh$

• Conservative force: A force for which the work done depends only on the initial and final positions.

Step-by-Step Guidance

1. Recall the definition of a conservative force and how it relates to work.

2. Write the formula for work done by gravity when lifting an object vertically.

3. Explain why the path taken does not affect the work done by gravity.

Try solving on your own before revealing the answer!