Objective: Calculate the acceleration of the two objects and the tension in the string.
For Acceleration
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \text{Force on heavier mass, } F_1 = m_1g | Weight of the heavier object (49.0 kg). |
| 2 | \text{Force on lighter mass, } F_2 = m_2g | Weight of the lighter object (24.0 kg). |
| 3 | \text{Net force, } F_{\text{net}} = F_1 – F_2 | The difference in weights provides the net force. |
| 4 | F_{\text{net}} = m_1g – m_2g | Substitute the values of F_1 and F_2. |
| 5 | F_{\text{net}} = (m_1 – m_2)g | Factor out g. |
| 6 | a = \frac{F_{\text{net}}}{m_1 + m_2} | Newton’s second law, acceleration equals net force divided by total mass. |
| 7 | a = \frac{(m_1 – m_2)g}{m_1 + m_2} | Combine steps 5 and 6. |
| 8 | a = \frac{(49.0\text{ kg} – 24.0\text{ kg})(9.8\text{ m/s}^2)}{49.0\text{ kg} + 24.0\text{ kg}} | Substitute the masses and gravitational acceleration. |
| 9 | a = \frac{25.0\text{ kg} \times 9.8\text{ m/s}^2}{73.0\text{ kg}} | Simplify the equation. |
| 10 | a = 3.356\text{ m/s}^2 | Calculate to find acceleration. |
Final answer for acceleration: \boxed{a = 3.356\text{ m/s}^2}
For Tension in the String
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | T – m_2g = m_2a | Newton’s second law for the lighter object. |
| 2 | T = m_2a + m_2g | Rearrange to solve for tension, T. |
| 3 | T = m_2(a + g) | Factor out m_2. |
| 4 | T = 24.0\text{ kg}(3.356\text{ m/s}^2 + 9.8\text{ m/s}^2) | Substitute the mass of the lighter object and calculated acceleration. |
| 5 | T = 24.0\text{ kg} \times 13.156\text{ m/s}^2 | Add a and g. |
| 6 | T = 315.744\text{ N} | Calculate to find tension. |
Final answer for tension: \boxed{T = 315.744\text{ N}}
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A skier with a mass of 58 kg glides up a snowy incline that forms an angle of 28 degrees with the horizontal. The skier initially moves at a speed of 7.2 m/s. After traveling a distance of 2.3 meters up the slope, the skier’s speed reduces to 3.8 m/s.
A car travels to right at constant velocity. The net force on the car is
A block starts from rest at the top of a 50° incline. The coefficient of kinetic friction between the block and the incline is 0.4. If the block reaches a velocity of 7 m/s at the bottom of the incline, what is the length of the incline?
Three blocks of masses m3 = 1.0, m2 = 2.0, and m1 = 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above.
The coefficient of kinetic friction is
3.36 m/s2, T = 316 N
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
| v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum mr^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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