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To solve the problem using principles of work and energy, we consider the system comprising the two masses. We know that one mass will move up while the other moves down, and we’re given the displacement d = 0.8 meters for each mass.

Assuming the 12 kg mass moves downward and the 9 kg mass moves upward, we need to find the common velocity v after they have moved 0.8 meters, taking into account that energy is conserved in this isolated system (ignoring air resistance and friction). Here’s the step-by-step analysis:

Step Derivation/Formula Reasoning
1 m_1 = 12 \, \text{kg}
m_2 = 9 \, \text{kg}
Define the masses where m_1 is the 12 kg mass and m_2 is the 9 kg mass.
2 g = 9.81 \, \text{m/s}^2 Acceleration due to gravity is g .
3 \Delta h = 0.8 \, \text{m} Both masses displace by 0.8 m, one moving up and the other down.
4 U_{\text{initial}} = 0 \, \text{J} Initial potential energy is set to zero for simplicity as we’re only interested in changes.
5 U_{\text{final}} = -m_1 g \Delta h + m_2 g \Delta h Final potential energy obtained by considering the gain in height by m_2 and loss in height by m_1 .
6 U_{\text{final}} = -12 \times 9.81 \times 0.8 + 9 \times 9.81 \times 0.8 Substitute the known values to find U_{\text{final}} .
7 U_{\text{final}} = -23.544 \, \text{J} Calculate the final potential energy.
8 KE_{\text{initial}} = 0 \, \text{J} Initial kinetic energy, assuming they start from rest.
9 KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v^2 Kinetic energy for the system of the two moving masses at the final state.
10 KE_{\text{final}} = \frac{1}{2} (12+9) v^2 Expression for the final kinetic energy of both masses.
11 KE_{\text{final}} = 10.5 v^2 Simplifying the expression for final kinetic energy.
12 KE_{\text{final}} = -U_{\text{final}} By conservation of energy, change in potential energy equals change in kinetic energy.
13 10.5 v^2 = 23.544 Substitute computed value of U_{\text{final}} .
14 v^2 = \frac{23.544}{10.5} Solve for v^2 .
15 v = \sqrt{\frac{23.544}{10.5}} Calculate v .
16 v \approx 1.49 \, \text{m/s} Final velocity of the masses after traveling 0.8 m.

This result indicates the common speed of both masses after they have moved 0.8 meters, based on the conservation of mechanical energy.

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1.49 m/s

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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)}
\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.

1. Start with the given measurement: \text{5 km}

2. 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}}

3. 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}

4. 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}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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