Physics 107: Midterm 2 Study Guidance

Q1. What is the maximum speed at which a 1200 kg car can negotiate a turn of radius 75 m if the coefficient of static friction between the tires and the road is 0.80?

Background

Topic: Circular Motion and Friction

This question tests your understanding of the forces involved in circular motion, specifically how friction provides the centripetal force needed for a car to safely make a turn without skidding.

Key Terms and Formulas

• Centripetal Force: The net force required to keep an object moving in a circle of radius $r$ at speed $v$ is $F_c = \frac{mv^2}{r}$.

• Static Friction: The maximum force of static friction is $F_{f,\text{max}} = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

• On a flat road, $N = mg$ (where $m$ is mass and $g$ is acceleration due to gravity).

Step-by-Step Guidance

1. Write the equation for the maximum static friction force: $F_{f,\text{max}} = \mu_s mg$.

2. Set the maximum static friction force equal to the required centripetal force for circular motion: $\mu_s mg = \frac{mv^2}{r}$.

3. Notice that the mass $m$ cancels out from both sides of the equation.

4. Rearrange the equation to solve for the maximum speed $v$ in terms of $\mu_s$, $g$, and $r$.

Try solving on your own before revealing the answer!

Final Answer: $v = \sqrt{\mu_s g r}$

Plugging in $\mu_s = 0.80$, $g = 9.8\ \text{m/s}^2$, and $r = 75\ \text{m}$ gives the maximum speed.

This formula shows that the maximum speed depends on the friction coefficient, gravity, and the turn radius.

Q2. A 0.5 kg mass is attached to a spring and oscillates horizontally with a period of 0.8 s. What is the spring constant?

Background

Topic: Simple Harmonic Motion (SHM)

This question tests your understanding of the relationship between mass, spring constant, and period for a mass-spring system undergoing simple harmonic motion.

Key Terms and Formulas

• Period of a Mass-Spring System: $T = 2\pi \sqrt{\frac{m}{k}}$

• $T$ is the period, $m$ is the mass, and $k$ is the spring constant.

Step-by-Step Guidance

1. Write the formula for the period: $T = 2\pi \sqrt{\frac{m}{k}}$.

2. Rearrange the formula to solve for the spring constant $k$.

3. Substitute the given values: $T = 0.8$ s, $m = 0.5$ kg.

4. Set up the equation for $k$ in terms of the known quantities, but do not calculate the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $k = \frac{4\pi^2 m}{T^2}$

Plug in $m = 0.5$ kg and $T = 0.8$ s to find the spring constant.

This formula comes from rearranging the period equation for a mass-spring system.

Q3. The acceleration due to gravity on Planet X is one-fourth that on Planet B. The radius of Planet X is one-half the radius of Planet B. What fraction of Planet B's mass is the mass of Planet X?

Background

Topic: Universal Gravitation

This question tests your understanding of how gravitational acceleration depends on a planet's mass and radius, using Newton's law of universal gravitation.

Key Terms and Formulas

• Gravitational Acceleration: $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the planet's mass, and $R$ is its radius.

Step-by-Step Guidance

1. Write the expression for gravitational acceleration on each planet: $g_X = \frac{G M_X}{R_X^2}$ and $g_B = \frac{G M_B}{R_B^2}$.

2. Express the given ratios: $g_X = \frac{1}{4} g_B$ and $R_X = \frac{1}{2} R_B$.

3. Set up the ratio $\frac{g_X}{g_B}$ using the formulas and substitute the given relationships.

4. Rearrange to solve for $\frac{M_X}{M_B}$, but do not compute the final fraction yet.

Try solving on your own before revealing the answer!

Final Answer: $\frac{M_X}{M_B} = \frac{1}{8}$

By substituting the ratios and simplifying, you find that Planet X has one-eighth the mass of Planet B.

Q4. A 5.0 kg crate is lifted by a person from ground level to the top of a 2.0 m shelf at a steady speed. What is the work done by the person on the crate?

Background

Topic: Work and Energy

This question tests your understanding of the concept of work done against gravity when lifting an object at constant speed.

Key Terms and Formulas

• Work Done Against Gravity: $W = Fd$, where $F$ is the force applied and $d$ is the displacement in the direction of the force.

• When lifting at constant speed, $F = mg$ (weight of the object).

Step-by-Step Guidance

1. Calculate the force needed to lift the crate: $F = mg$.

2. Identify the vertical displacement: $d = 2.0$ m.

3. Set up the work equation: $W = Fd$.

4. Substitute the values for $F$ and $d$, but do not multiply them out yet.

Try solving on your own before revealing the answer!

Final Answer: $W = (5.0\ \text{kg})(9.8\ \text{m/s}^2)(2.0\ \text{m}) = 98\ \text{J}$

The work done equals the weight of the crate times the height lifted.

Q5. A 0.5 kg block is moving at 5.0 m/s and slides across a horizontal surface. If the coefficient of kinetic friction is 0.10, how far does the block slide before coming to rest?

Background

Topic: Work-Energy Principle and Friction

This question tests your ability to apply the work-energy theorem and calculate the effect of friction on a moving object.

Key Terms and Formulas

• Kinetic Energy: $KE = \frac{1}{2}mv^2$

• Work Done by Friction: $W_f = -f_k d$, where $f_k = \mu_k N$ and $N = mg$ on a horizontal surface.

• Work-Energy Theorem: $\Delta KE = W_{\text{friction}}$

Step-by-Step Guidance

1. Calculate the initial kinetic energy: $KE_i = \frac{1}{2}mv^2$.

2. Write the expression for the frictional force: $f_k = \mu_k mg$.

3. Set up the work-energy theorem: $0 - KE_i = -f_k d$ (since the block comes to rest).

4. Rearrange to solve for $d$ in terms of the given quantities, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $d = \frac{\frac{1}{2}mv^2}{\mu_k mg}$

Plug in $m = 0.5$ kg, $v = 5.0$ m/s, and $\mu_k = 0.10$ to find the distance.

This formula comes from equating the initial kinetic energy to the work done by friction.