Answer:
a. 0.41 m
b. 5.72 m/s
c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.
ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.
d. 5.72 m/s
Explanation:
a. Use energy to find how high the penny goes above your hand before stopping.
Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.
E = E'
U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)
So, U + K = U' + K'
mgh + 1/2mv² = mgh' + 1/2mv'²
substituting the values of the variables into the equation, we have
mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²
0 + 1/2m(8.1225 m²/s²) = mgh' + 0
m(4.06125 m²/s²) = mgh'
h' = 4.06125 m²/s² ÷ g
h' = 4.06125 m²/s² ÷ 9.8 m/s²
h' = 0.41 m
(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.
Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E' equals the total mechanical energy on the ground, E"
E' = E"
U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,
So, U' + K' = U' + K'
mgh" + 1/2mv'² = mgh₁ + 1/2mv"²
substituting the values of the variables into the equation, we have
mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²
1.67mg + 0 = 0 + 1/2mv"²
1.67mg = 1/2mv"²
1.67g = 1/2v"²
v"² = 2(1.67g)
v" = √[2(1.67g)]
v" = √[2(1.67 m × 9.8 m/s²)]
v" = √[2(16.366 m²/s²)]
v" = √[32.732 m²/s²)]
v" = 5.72 m/s
(c) Explain your choice of reference level for parts (a) and (b).
i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.
ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.
(d) Choose a different reference level and repeat part (b)
Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E' equals the total mechanical energy on the ground, E"
E' = E"
U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,
So, U' + K' = U' + K'
mgh' + 1/2mv'² = mgh₂ + 1/2mv"²
substituting the values of the variables into the equation, we have
mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²
0.41mg + 0 = -1.26 mg + 1/2mv"²
0.41mg + 1.26mg = 1/2mv"²
1.67mg = 1/2mv"²
1.67g = 1/2v"²
v"² = 2(1.67g)
v" = √[2(1.67g)]
v" = √[2(1.67 m × 9.8 m/s²)]
v" = √[2(16.366 m²/s²)]
v" = √[32.732 m²/s²)]
v" = 5.72 m/s
Sean and Tommy are debating the positive and negative impacts of technology on environmental quality. Tommy is listing the negative impacts. Which of the following would help his argument?
A. Solar panels have become more effective due to technological improvements.
B. Technology has helped make recycling to be more efficient.
C. It has created new ways to clean up oil spills.
D. Factories contribute to air and water pollution
A car takes a full round of journey in a roundabout with constant speed, as the driver got confused with the route. Can we consider it as a uniform motion? Why?
Answer: The given statement is True
Explanation:
A uniform motion is defined as the motion where an object is moving at a constant speed.
A non-uniform motion is defined as the motion where an object keeps changing its position and does not move at a constant speed.
We are given:
A car takes a full round of journey in a roundabout with constant speed
As the speed remains constant in a circular path, it is considered a uniform motion.
Hence, the given statement is True
A horse pulls a wagon with a force of 200 N for a distance of 80 m. How much work
does the horse do?
Answer:
w=f×s
w = 16000 J, hope this helps
En una sala de juntas hay mesas, sillas y otras personas. ¿Cuál de ellas tienen temperaturas
a) menores, b) mayores y d) iguales que la del aire?
Answer:
table and chair
Explanation:
In a meeting room there are tables, chairs, and other people. Which of them have temperatures
a) smaller, b) bigger and d) equal to that of air?
the temperature of tables and chairs is same as air.
Find the emitted power per square meter and wavelength of peak intensity for a 3000 K object that emits thermal radiation.
Answer:
power per square meter = 4.593 × 10^(6) W/m²
Wavelength of peak intensity = 9.67 × 10^(-7) m
Explanation:
From Stefan-Boltzmann law, total emitted power per square meter is given as;
P/A = eσT⁴
where;
P is power
A is surface area
σ = Stefan-Boltzmann constant = 5.67 × 10^(-8) W/m².k⁴
T = temperature of the body = 3000 K
e = emissivity of the substance (for ideal radiation, it has a value = 1)
Thus, Plugging in the relevant values we have;
P/A = 1 × 5.67 × 10^(-8) × (3000)^(4)
P/A = 4.593 × 10^(6) W/m²
Let's find the wavelength of peak intensity.
From wiens displacement law, we know that;
λ_m × T = b
where;
λ_m = maximum wavelength
T = Temperature
b is Wien's displacement constant = 2.9 × 10^(−3) m/K
thus;
λ_m = b/T = (2.9 × 10^(−3))/3000 = 9.67 × 10^(-7) m
13) Un móvil A parte de una ciudad a las 12 horas, con una velocidad de 40 Km/h. 2 horas después parte otro con una velocidad de 60 Km/h. Averiguar a qué hora se encuentran y a que distancia de la ciudad
Answer:
¿Podrías poner la pregunta en inglés por favor?
Explanation:
Select the correct answer.
Which quantity is a vector quantity?
ОА. .
acceleration
OB.
mass
OC.
speed
O D.
volume
Reset
Next
Answer:
acceleration is the vector quantity because it depends on particular direction and has magnitude
A machine has mechanical advantage 2.What does that mean
Answer:
Mechanical advantage is the measure of the force amplification achieved by using a tool , mechanical device or machine . The machine preserve the input power and supply trade off force. against movement to obtain a desired amplification in the output force .
Answer:
The ratio of load overcome by the machine to the effort applied is called mechanical advantage.
Why is a person not a good blackbody radiator?
O A. A person emits only visible light.
OB. A person emits only infrared radiation.
O C. A person absorbs most of the light that hits him or her.
O D. A person reflects little of the light that hits him or her.
Answer:
O C. A person absorbs most of the light that hits him or her.
Explanation:
Answer:
Option D hope it's helpful mark me as brainlistExplain: What happens to the velocity of a stream as the size of the sediment increases?
Answer:
Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.
On which planet would your weight be the most and the least?
a) Jupiter and Mercury
b. Jupiter and neptune
c. Saturn and Neptune
d. Saturn and uranus
Answer:
Jupiter and neptune
Explanation:
ball of mass M kg is dropped from rest off a high building and strikes the ground 10 seconds later. Calculate the height of the building assuming that upwards is positive in this coordinate system
Answer:
h = 1/2gt^2
Is the formula for the height of a free falling object.
Put in the numbers:
h = 1/2*9.8 (gravitational constant) * 10^2
h=490 meters.
Please note that the gravitational constant is approximate, you could also use a more specific value or 10
A HIGH SPEED TRAIN IS 180M LONG AND IT IS TRAVELLING AT 50M/S.HOW LONG WILL IT TAKE TO PASS A PERSON STANDING AT A LEVEL CROSSING?
B-HOW LONG WILL IT TAKE TO PASS COMPLETELY THROUGH A STATION WHOSE PLATFORMS ARE 220M IN LENGTH?
Answer:
a. Time = 3.6 seconds
b. Time = 4.4 seconds
Explanation:
Given the following data;
Distance = 180 m
Speed = 50 m/s
a. To find the time;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
[tex]Speed = \frac{distance}{time}[/tex]
Making time the subject of formula, we have;
[tex]Time = \frac{distance}{speed}[/tex]
Substituting into the equation, we have;
[tex]Time = \frac{180}{50}[/tex]
Time = 3.6 seconds
b. Distance = 220 meters
Speed = 50 m/s
To find the time;
[tex]Time = \frac{distance}{speed}[/tex]
Substituting into the equation, we have;
[tex]Time = \frac{220}{50}[/tex]
Time = 4.4 seconds
Melanie gets into an accident on the highway that sends her to the hospital for three weeks with multiple broken bones. Her hospital bill totals over $32,000, but she discovers that the woman who hit her only has $25,000 worth of liability insurance.
The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is received with a time delay of 0.50s after the original sound is emitted.
Calculate the distance between submarine and the other object
Answer:
d = 375 m
Explanation:
The speed of sound is constant in any medium, therefore we can use the uniform motion relationships
v = x / t
x = v t
In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance
x = 2d
2d = v t
d = v t/2
let's calculate
d = 1500 0.5 / 2
d = 375 m
if the action force is 100N what will be the reaction force
Answer:
HONORS PHYSICS
Introduction
Matter & Energy
Math Review
Kinematics
Defining Motion
Graphing Motion
Kinematic Equations
Free Fall
Projectile Motion
Relative Velocity
Dynamics
Newton's 1st Law
Free Body Diagrams
Newton's 2nd Law
Static Equilibrium
Newton's 3rd Law
Friction
Ramps and Inclines
Atwood Machines
Momentum
Impulse & Momentum
Conservation Laws
Types of Collisions
Center of Mass
UCM & Gravity
Uniform Circular Motion
Gravity
Kepler's Laws
Rotational Motion
Rotational Kinematics
Torque
Angular Momentum
Rotational KE
Work, Energy & Power
Work
Hooke's Law
Power
Energy
Conservation of Energy
Fluid Mechanics
Density
Pressure
Buoyancy
Pascal's Principle
Fluid Continuity
Bernoulli's Principle
Thermal Physics
Temperature
Thermal Expansion
Heat
Phase Changes
Ideal Gas Law
Thermodynamics
Electrostatics
Electric Charges
Coulomb's Law
Electric Fields
Potential Difference
Capacitors
Current Electricity
Electric Current
Resistance
Ohm's Law
Circuits
Electric Meters
Circuit Analysis
Magnetism
Magnetic Fields
The Compass
Electromagnetism
Microelectronics
Silicon
P-N Junctions
Transistors
Digital Logic
Processing
Integration
Waves & Sound
Wave Characteristics
Wave Equation
Sound
Interference
Doppler Effect
Optics
Reflection
Refraction
Diffraction
EM Spectrum
Modern Physics
Wave-Particle Duality
Models of the Atom
M-E Equivalence
The Standard Model
Relativity
DYNAMICS
Newton's 1st Law
Free Body Diagrams
Newton's 2nd Law
Static Equilibrium
Newton's 3rd Law
Friction
Ramps and Inclines
Atwood Machines
Silly Beagle
Newtons's 3rd Law of Motion
Newton’s 3rd Law of Motion, commonly referred to as the Law of Action and Reaction, describes the phenomena by which all forces come in pairs. If Object 1 exerts a force on Object 2, then Object 2 must exert a force back on Object 1. Moreover, the force of Object 1 on Object 2 is equal in magnitude, or size, but opposite in direction to the force of Object 2 on Object 1. Written mathematically:
Newton's 3rd Law Equation
This has many implications, some of which aren’t immediately obvious. For example, if you punch the wall with your fist with a force of 100N, the wall imparts a force back on your fist of 100N (which is why it hurts!). Or try this. Push on the corner of your desk with your palm for a few seconds. Now look at your palm... see the indentation? That’s because the corner of the desk pushed back on your palm.
running tiger
Although this law surrounds your actions everyday, often times you may not even realize its effects. To run forward, a cat pushes with its legs backward on the ground, and the ground pushes the cat forward. How do you swim? If you want to swim forwards, which way do you push on the water? Backwards, that’s right. As you push backwards on the water, the reactionary force, the water pushing you, propels you forward. How do you jump up in the air? You push down on the ground, and it’s the reactionary force of the ground pushing on you that accelerates you skyward!
As you can see, then, forces always come in pairs. These pairs are known as action-reaction pairs. What are the action-reaction force pairs for a girl kicking a soccer ball? The girl’s foot applies a force on the ball, and the ball applies an equal and opposite force on the girl’s foot.
How does a rocket ship maneuver in space? The rocket propels hot expanding gas particles outward, so the gas particles in return push the rocket forward. Newton’s 3rd Law even applies to gravity. The Earth exerts a gravitational force on you (downward). You, therefore, must apply a gravitational force upward on the Earth!
A gas occupies 2 m^3 at 27°C at a pressure of 1 atmosphere. At a pressure of 2 atmospheres it occupies a volume of 1 m^2. What is its temperature at this new volume and pressure?
Answer:
27°C
Explanation:
We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 27 °C
Initial temperature (T₁) = 27 °C + 273
Initial temperature (T₁) = 300 K
Next, we shall determine the final temperature of the gas. This can be obtained as follow:
Initial volume (V₁) = 2 m³
Initial temperature (T₁) = 300 K
Initial pressure (P₁) = 1 atm
Final pressure (P₂) = 2 atm
Final volume (V₂) = 1 m³
Final temperature (T₂) =?
P₁V₁/T₁ = P₂V₂/T₂
1 × 2 / 300 = 2 × 1 / T₂
2/300 = 2/T₂
1/150 = 2/T₂
Cross multiply
T₂ = 150 × 2
T₂ = 300 K
Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 300 K
T(°C) = 300 – 273
T(°C) = 27°C
Thus, the final temperature is 27°C
The auroras occur in the
C.
a. troposphere
b. stratosphere
mesophere
d. ionosphere
Answer:
Ionosphere
Explanation:
The thermosphere reaches 600 kilometres just above mesosphere and begins immediately above the mesosphere. This layer is where the aurora and satellites appear.
The ionosphere is the comprehensive career of the mesosphere because most of the thermosphere, located 80–400 kilometres just above ground atmosphere.
Auroras — magnificent flowing streaks of light seen in the night sky – appear in this location.
help....................
Em um fio condutor uma carga de 6.000 C atravessa uma secção transversal em 5 minutos. Determinando-se a corrente no fio, encontraremos o valor de?
Answer:
I = 20 A
Explanation:
The question says that, "A load of 6,000 C is conducted through a cross section in 5 minutes. Determining-if a current is not correct, we will find the value of?"
We have,
Charge, q = 6,000 C
Time, t = 5 minutes = 300 s
We need to find the current. We know that, the charge flowing per unit time is equal to current. So,
[tex]I=\dfrac{q}{t}\\\\I=\dfrac{6000}{300}\\\\I=20\ A[/tex]
So, the current flowing through the circuit is 20 A.
When magma flows on the surface on the surface, it is already called lava
TRUE OR FALSE
Answer:
True
Explanation:
I guess you made a mistake on question.
but I understood what you wanted to say.
Hope this helps... :)
In a bicycle dynamo,does 1. The permanent magnet surrounds a conducting coil 2. The conducting coil rotates when the rear wheel of the bicycle rotates. 3.The electricity is generated in the permanent magnet 4.When the rear wheel of the bicycle turns fast, the brightness of the light increases
Explanation:
3: the electricity is generated in the permanent magnet
distance= 10km due West in 1hour calculate the velocity
Answer:
Velocity = distance / time
V = 10/1
V = 10km/h
Answer:10km/h or 2.77m/s.
Explanation:
Distance =10km
Time =1h
Velocity =10/1 =10km/h
Or,
Distance =10km =10000m
Time =1h =60min = 3600s
Velocity =10000/3600 =2.77m/s
sulfur and oxygen can react to form both sulfur dioxide and sulfur trioxide in sulfur dioxide there are 32.06 grams of sulfur and 32 grams of oxygen in sulfur dioxide there are 32.06 grams of sulfur are combined with 48 grams of oxygen
a. what is the ratio of the weights of oxygen that combine with 32.06 g of sulfur ?
b. How do these data illustrate the law of multiple proportions?
Answer:
a. 2:3
b. The data illustrates the law of multiple proportions by showing that the the masses of oxygen that reacts with a fixed mass of sulfur are in a ratio of small whole numbers
Explanation:
The weight of oxygen that combines with 32.06 grams of sulfur in sulfur dioxide = 32 grams
The weight of oxygen that combines with 32.06 grams of sulfur in sulfur trioxide = 48 grams
a. The ration of the weights of oxygen that combine with 32.06 g of sulfur = 32:48 = 2:3
b. The law of multiple proportions states that when two elements are able to interact chemically to form more than one compound, then the (different) weights of one of the element that combines with a fixed weight of the other element are in small whole number ratios
The data demonstrates the law of multiple proportions by showing that the ratios of the weights of oxygen that combine with a fixed weight of sulfur to form sulfur dioxide and sulfur trioxide is in the ratio of 2 to 3 which are small whole number ratios
The earth's orbital is oval in shape. Explain how the magnitude of the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure.
The gravitational force does change believe it or not, but the explaination for this is because the earths orbit is an oval (or a not circle) the closer it nears its self to the sun.
A student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g = 9.8 m/s^2). What is the ball speed just before hitting the ground?
Answer:
Explanation:
Here's what we know because it was given to us:
a = -9.8 m/s/s and
time = 3.32 seconds
Here's what we know because we rock physics:
v₀ = 0 (because the object was held still before it was dropped).
Here's the equation that ties all that info together in a single one-dimensional equation:
v = v₀ + at
Filling in and solving for v:
v = 0 + (-9.8)(3.32) and
v = -33m/s
The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)
A runner has a speed of 5 m/s and a mass of 130 kg. What is his kinetic
energy?
O A. 1625 J
B. 3250 J
C. 875 J
D. 325 J
Neha and Reha are playing see-saw.Neha is sitting 60cm away from the fulcrum and Reha is sitting 40cm away from the fulcrum.Calculate the effort that Reha should apply to lift Neha.The weight of Neha is 360N.
Answer:
Effort = 540 Newton
Explanation:
Given the following data;
Load arm = 60 cm
Effort arm = 40 cm
Load = 360 N
Conversion:
100 cm = 1 meters
40 cm = 40/100 = 0.4 meters
60 cm = 60/100 = 0.6 meters
To calculate the effort that Reha should apply to lift Neha, we would use the expression;
Effort * effort arm = load * load arm
Substituting into the expression, we have;
Effort * 0.4 = 360 * 0.6
Effort * 0.4 = 216
Effort = 216/0.4
Effort = 540 Newton
A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block
of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2.
After the collision the bullet becomes embedded in the block.
A) What is the net momentum of the bullet-block system before the collision?
B) Find the total energy of the bullet-block system before the collision?
C) What is the speed of the bullet-block system after the collision?
D) *Find the total energy of the bullet-block system after the collision?
E) *How much work must be done to stop the bullet-block system?
F) *Find the maximum traveled distance of the bullet-block after the collision?
Answer:
a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.
b) The initial translational kinetic energy of the bullet before the collision is 625 joules.
c) The final speed of the bullet-block system after the collision is 7.143 meters per second.
d) The total energy of the bullet-block system after the collision is 89.289 joules.
e) 89.289 joules must be done to stop the bullet-block system.
f) The bullet-block system will travel 13.007 meters before stopping.
Explanation:
a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by the initial momentum of the bullet:
[tex]p = m\cdot v_{o}[/tex] (1)
Where:
[tex]p[/tex] - Net momentum, in kilogram-meters per second.
[tex]m[/tex] - Mass of the bullet, in kilograms.
[tex]v_{o}[/tex] - Initial speed of the bullet, in meters per second.
If we know that [tex]m = 0.5\,kg[/tex] and [tex]v_{o} = 50\,\frac{m}{s}[/tex], then the net momentum of the bullet-block system before the collision is:
[tex]p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)[/tex]
[tex]p = 25\,\frac{kg\cdot m}{s}[/tex]
The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.
b) The total energy of the bullet before the collision is its initial translational kinetic energy ([tex]K[/tex]), in joules:
[tex]K = \frac{1}{2}\cdot m \cdot v_{o}^{2}[/tex] (2)
[tex]K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 625\,J[/tex]
The initial translational kinetic energy of the bullet before the collision is 625 joules.
c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:
[tex]v_{f} = \frac{m\cdot v_{o}}{m+M}[/tex] (3)
Where:
[tex]v_{f}[/tex] - Final speed, in meters per second.
[tex]M[/tex] - Mass of the block, in kilograms.
If we know that [tex]m = 0.5\,kg[/tex], [tex]v_{o} = 50\,\frac{m}{s}[/tex] and [tex]M = 3\,kg[/tex], then the final speed of the bullet-block system is:
[tex]v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)[/tex]
[tex]v_{f} = 7.143\,\frac{m}{s}[/tex]
The final speed of the bullet-block system after the collision is 7.143 meters per second.
d) The total energy of the bullet-block system after the collision is the translational kinetic energy of the system ([tex]K[/tex]), in joules, is:
[tex]K = \frac{1}{2}\cdot (m + M)\cdot v_{f}^{2}[/tex] (4)
[tex]K = \frac{1}{2}\cdot (0.5\,kg + 3\,kg)\cdot \left(7.143\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 89.289\,J[/tex]
The total energy of the bullet-block system after the collision is 89.289 joules.
e) By Work-Energy Theorem, magnitude of the work done by friction must be equal to the magnitude of the translational kinetic energy of the system. Hence, 89.289 joules must be done to stop the bullet-block system.
f) The maximum travelled distance of the bullet-block after the collision can be determined by means of Work-Energy Theorem and definition of work:
[tex]W = \mu_{k}\cdot (m+M)\cdot g\cdot s[/tex] (5)
Where:
[tex]W[/tex] - Work done by friction, in joules.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
If we know that [tex]m = 0.5\,kg[/tex], [tex]M = 3\,kg[/tex], [tex]\mu_{k} = 0.2[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]W = 89.289\,J[/tex], then the travelled distance of the bullet-block system is:
[tex]s = \frac{W}{\mu_{k}\cdot (m+M)\cdot g}[/tex]
[tex]s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]s = 13.007\,m[/tex]
The bullet-block system will travel 13.007 meters before stopping.
Convert:
1) 2kg into gram
2) 5200m into km
3) 20cm into m