Physics Exam Study Guide: Work, Energy, and Momentum

Q1. A 100 kg box slides down a frictionless incline. If the gravitational potential energy at the top is 1500 J, what is the kinetic energy at the bottom?

Background

Topic: Conservation of Energy

This question tests your understanding of how energy transforms from potential to kinetic in the absence of friction.

Key Terms and Formulas:

• Gravitational Potential Energy: $U = mgh$

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

• Conservation of Mechanical Energy: $U_{\text{top}} + K_{\text{top}} = U_{\text{bottom}} + K_{\text{bottom}}$

Step-by-Step Guidance

1. At the top, the box has gravitational potential energy ($U_{\text{top}} = 1500$ J) and zero kinetic energy (since it starts from rest).

2. At the bottom, all potential energy is converted to kinetic energy (since the incline is frictionless).

3. Set up the conservation of energy equation: $U_{\text{top}} = K_{\text{bottom}}$.

4. Substitute the given value for $U_{\text{top}}$ to find $K_{\text{bottom}}$.

Try solving on your own before revealing the answer!

Q2. A 20 N horizontal force is applied to a 4.0 kg box (friction negligible), moving it 3.0 meters. What is the work done by the force?

Background

Topic: Work and Energy

This question tests your ability to calculate work done by a constant force over a distance.

Key Terms and Formulas:

• Work: $W = Fd\cos\theta$

• Where $F$ is force, $d$ is displacement, and $\theta$ is the angle between force and displacement.

Step-by-Step Guidance

1. Identify the values: $F = 20$ N, $d = 3.0$ m, $\theta = 0^\circ$ (force and displacement are in the same direction).

2. Recall that $\cos 0^\circ = 1$.

3. Plug the values into the work formula: $W = Fd\cos\theta$.

4. Set up the multiplication for the final calculation.

Try solving on your own before revealing the answer!

Q3. What initial speed is needed for an 8000 kg satellite to escape Earth's gravity (reach infinite distance), neglecting air resistance?

Background

Topic: Gravitational Potential Energy and Escape Velocity

This question tests your understanding of escape velocity and energy conservation in gravitational fields.

Key Terms and Formulas:

• Escape Velocity: $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

• $G$ = universal gravitational constant ($6.67 \times 10^{-11}$ N·m$^2$/kg$^2$)

• $M$ = mass of Earth ($5.97 \times 10^{24}$ kg)

• $R$ = radius of Earth ($6.37 \times 10^6$ m)

Step-by-Step Guidance

1. Write the formula for escape velocity: $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$.

2. Identify the values for $G$, $M$, and $R$.

3. Substitute the values into the formula.

4. Set up the calculation under the square root, but do not compute the final value yet.

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Q4. A 1800 kg car moving at 20 m/s hits a spring (k = $2.0 \times 10^6$ N/m). What is the maximum compression of the spring?

Background

Topic: Conservation of Energy (Kinetic and Elastic Potential Energy)

This question tests your ability to apply energy conservation to a system involving kinetic and elastic potential energy.

Key Terms and Formulas:

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

• Elastic Potential Energy: $U_s = \frac{1}{2}kx^2$

• Energy Conservation: $K_{\text{initial}} = U_{s,\text{max}}$

Step-by-Step Guidance

1. Calculate the initial kinetic energy of the car using $K = \frac{1}{2}mv^2$.

2. Set this equal to the maximum elastic potential energy stored in the spring: $\frac{1}{2}kx^2$.

3. Set up the equation: $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$.

4. Rearrange to solve for $x$ (maximum compression): $x = \sqrt{\frac{mv^2}{k}}$.

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Q5. A 2.0 kg ball is moving at 4.0 m/s WEST. What is its momentum?

Background

Topic: Linear Momentum

This question tests your understanding of how to calculate the momentum of an object.

Key Terms and Formulas:

• Momentum: $p = mv$

• Direction matters: include the direction (WEST) in your answer.

Step-by-Step Guidance

1. Identify the mass ($m = 2.0$ kg) and velocity ($v = 4.0$ m/s WEST).

2. Plug these values into the momentum formula: $p = mv$.

3. Set up the multiplication for the final calculation, remembering to include the direction.

Try solving on your own before revealing the answer!

Q6. A 3.0 kg object (right, 4.0 m/s) collides inelastically with a 6.0 kg object (left, 2.0 m/s). What is the total kinetic energy after the collision?

Background

Topic: Collisions and Conservation of Momentum

This question tests your ability to analyze perfectly inelastic collisions and calculate kinetic energy after the collision.

Key Terms and Formulas:

• Momentum Conservation: $m_1v_1 + m_2v_2 = (m_1 + m_2)V_f$

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

• Perfectly Inelastic: Objects stick together after collision.

Step-by-Step Guidance

1. Assign directions: right is positive, left is negative.

2. Calculate the total initial momentum: $p_{\text{initial}} = m_1v_1 + m_2v_2$.

3. Set $p_{\text{initial}} = (m_1 + m_2)V_f$ and solve for $V_f$ (final velocity).

4. Calculate the total kinetic energy after the collision using $K = \frac{1}{2}(m_1 + m_2)V_f^2$.

Try solving on your own before revealing the answer!