Part a: Calculate the acceleration of the system
Step | Formula Derivation | Reasoning |
---|---|---|
1 | F_{\text{net,A}} = m_A \cdot a | Net force on mass A equals mass times acceleration. |
2 | F_{\text{net,B}} = m_B \cdot a | Net force on mass B equals mass times acceleration. |
3 | F_{\text{net,A}} = T – m_A \cdot g | Tension upwards minus weight of A downwards. |
4 | F_{\text{net,B}} = m_B \cdot g – T | Weight of B downwards minus tension upwards. |
5 | m_A \cdot a = T – m_A \cdot g | Substitute step 1 into step 3. |
6 | m_B \cdot a = m_B \cdot g – T | Substitute step 2 into step 4. |
7 | m_A \cdot a + m_B \cdot a = m_B \cdot g – m_A \cdot g | Add step 5 and step 6 equations. |
8 | a = \frac{(m_B – m_A) \cdot g}{m_A + m_B} | Solve for acceleration a. |
Use the given number from the problem.
Step | Formula Derivation | Reasoning |
---|---|---|
9 | a = \frac{(2.4 – 3.2) \cdot 9.8}{3.2 + 2.4} | Plug in known values. |
10 | a = \frac{-0.8 \cdot 9.8}{5.6} | Simplify the numerator and denominator. |
11 | a = -1.4 , \text{m/s}^2 | Calculate the acceleration. |
\boxed{a = -1.4 , \text{m/s}^2} Negative sign indicates downward direction.
Part b: Calculate the tension in the string
Step | Formula Derivation | Reasoning |
---|---|---|
1 | T = m_A \cdot (g + a) | Tension equals mass A times (gravity plus acceleration). |
Using the previously calculated acceleration:
Step | Formula Derivation | Reasoning |
---|---|---|
2 | T = 3.2 \cdot (9.8 – 1.4) | Plug in known values for mass and acceleration |
3 | T = 3.2 \cdot 8.4 | Simplify the terms in the parentheses. |
4 | T = 26.88 , \text{N} | Calculate the tension. |
Part c: Calculate the final speed of mass A before it hits the ground
Step | Formula Derivation | Reasoning |
---|---|---|
1 | v = \sqrt{2 \cdot} | a |
2 | v = \sqrt{2 \cdot 1.4 \cdot 0.5} | Plug in the magnitude of a and s = 0.5m. |
3 | v = \sqrt{1.4} | Calculate the expression under the square root. |
4 | v = 1.18 , \text{m/s} | Find the square root to get the final velocity. |
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You are standing on a bathroom scale in an elevator. The elevator starts from rest on the first floor and accelerates up to the third floor, 12 meters above, in a time of 6 seconds. The scale reads 800N. What is the mass of the person?
There are two cables that lift an elevator, each with a force of 10,000 N. The 1,000 kg elevator is lifted from the first floor and accelerates over 10 m until it reaches its top speed of 6 m/s. What is the mass of the people in the elevator?
What is weight of a person who has a mass of 75 kg?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
For linear motion the term “inertia” refers to the same physical concept of
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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} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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