Answer:
μ = 0.15
Explanation:
Let's start by using Hooke's law to find the force applied to the block
F = k x
F = 87.0 0.065
F = 5.655 N
Now we use the translational equilibrium relation since the block has no acceleration
∑ F = 0
F -fr = 0
F = fr
the expression for the friction force is
fr = μ N
if we write Newton's second law for the y-axis
N -W = 0
N = W = mg
we substitute
F = μ mg
μ = F / mg
μ = [tex]\frac{5.655}{3.8 \ 9.8}[/tex]
μ = 0.15
What is the strength of the magnetic field a distance 4.4 mm above the center of a circular loop of radius 0.8 mm and current 474.1 A
Answer:
B = 0.118 T
Explanation:
From Biot-Savart Law:
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where,
B = strength of magnetic field = ?
μ₀ = 4π x 10⁻⁷ Tm/A
I = current enclosed = 474.1 A
r = radius = 0.8 mm = 8 x 10⁻⁴ m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]
B = 0.118 T
A wave pulse travels along a stretched string at a speed of 200 cm/s. What will be the speed if:
a. The string's tension is doubled?
b. The string's mass is quadrupled (but its length is unchanged)?
c. The string's length is quadrupled (but its mass is unchanged)?
d. The string's mass and length are both quadrupled?
Answer:
a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s
Explanation:
The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.
So, v = √(T/μ)
v = √(T/m/l)
v = √(Tl/m)
a. The string's tension is doubled?
If the tension is doubled, T' = 2T the new speed is
v' = √(T'l/m)
v' = √(2Tl/m)
v' = √2(√Tl/m)
v' = √2v
v' = √2 × 200 cm/s
v' = 282.84 cm/s
b. The string's mass is quadrupled (but its length is unchanged)?
If the mass is quadrupled, m' = 4m the new speed is
v' = √(Tl/m')
v' = √(Tl/4m)
v' = (1/√4)(√Tl/m)
v' = v/2
v' = 200/2 cm/s
v' = 100 cm/s
c. The string's length is quadrupled (but its mass is unchanged)?
If the length is quadrupled, l' = 4l the new speed is
v' = √(Tl'/m)
v' = √(T(4l)/m)
v' = √4)(√Tl/m)
v' = 2v
v' = 200 × 2 cm/s
v' = 400 cm/s
d. The string's mass and length are both quadrupled?
If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is
v' = √(Tl'/m')
v' = √(T(4l)/4m)
v' = √(Tl/m)
v' = v
v' = 200 cm/s
An audience of 2250 fills a concert hall of volume 32000 m^3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?
suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along
Answer:
Pressure is more in the open container than the closed one.
Explanation:
The pressure due to the fluid at a depth is given by
Pressure = depth x density of fluid x gravity
So, when the container is open, the atmospheric pressure is also add up but when the container is closed only the pressure due to the fluid is there.
So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.
hen the container is closed only the pressure due to the fluid is there.
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
A wheel accelerates so that it's angular speed increases uniformly from 150 rads/s to 580 rads/s in 16 revolutions.Cakcjlate its angular acceleration.
Answer:
A = 26.875 rad/s²
Explanation:
Given the following data;
Initial angular speed, Uw = 150 rads/s.
Final angular speed, Vw = 580 rads/s.
Time = 16 seconds.
To calculate the angular acceleration;
From kinematics equation;
At = Vw - Uw
Where;
A is the angular acceleration.t is the timeVw is the final angular speed.Uw is the initial angular speed.Substituting into the formula, we have;
A*16 = 580 - 150
16A = 430
A = 430/16
A = 26.875 rad/s²
If earth is compressed to the volume of moon, its acceleration due to gravity
* i. decreases
ii. remains same as before
iii. increases
iv. none of these
Answer:
increase
Explanation:
hope it will help you:)
Mary needs to row her boat across a 160 m-wide river that is flowing to the east at a speed of 1.5 m/s. Mary can row with a speed of 3.6 m/s. If Mary points her boat due north, how far from her intended landing spot will she be when she reaches the opposite shore? What is her speed with respect to the shore?
Answer: 66.67 m, 44.44 s
Explanation:
Given
Velocity of flow is [tex]u=1.5\ m/s[/tex]
Mary can row with speed [tex]v=3.6\ m/s[/tex]
Width of the river [tex]y=160\ m[/tex]
Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction
time taken to cross river
[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]
Flow will drift Mary by
[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]
Velocity w.r.t shore is
[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]
A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.
"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.
The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)
Using Newton's second law, we set up the following equations.
• p = 100 N
∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0
∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0
• P = 350 N
∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0
∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0
(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)
Solve for n and N :
n = p sin(θ) + mg cos(θ)
N = P sin(θ) - mg cos(θ)
Substitute these into the corresponding equations containing µ, and solve for µ :
µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))
µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
Next, you would set these equal and solve for m :
(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
...
Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with
m ≈ 36.5 kg
µ ≈ 0.256
The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.
From the given,
The force that opposes the crate by sliding is P = 100N
In X-axis, the sum of forces is zero.
ΣF = 0
Pcosθ - mgsinθ-Ff = 0
Ff = Pcosθ - mgsinθ
In Y-axis
Psinθ - mgcosθ - N = 0
N = Psinθ-mgcosθ
Frictional force, Ff = μN, μ is the coefficient of friction
Ff = μN
Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0
μ = mgsin30-Pcos30/Psin30+mgcos30 ------1
The block is sliding with the horizontal force, F = 350N
X-axis
P₂cosθ - mgsinθ-Ff = 0
Y-axis
P₂sinθ - mgcosθ - N = 0
N = P₂sinθ-mgcosθ
μ = P₂cos30-mgsin30/P₂sin30-mgcos30 -----2
Equate equations 1 and 2
mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30
4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49
41.7m² + 123m - 1.516×10⁴ = 0
-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)
83.4m² - 2207m -3.03×10⁴ = 0
m= 36.4 kg
Hence, the mass of the crate is 36.4 Kg.
Substitute the value of m in equation 1,
μ = 4.905(36.4) - 86.6 / 50 + 8.49
μ = 0.256
Thus, the coefficient of static friction is 0.256.
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The following contribute to accidents when a teen driver has other teens as passengers
Answer:
When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
What are the factors contribute to accidents when a teen driver has other teens as passengers?When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.
Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.
Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
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State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.
Answer:
non biodegradable
Explanation:
It is non biodegradable because plastic cannot dispose off easily ..
A 200-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 30 ft. If the silo is 60 ft high and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top
Answer:
17.07 kJ
Explanation:
The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU
W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft
Since W = mgΔy, we substitute the values of the variables into the equation.
So,
W = mgΔy
W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft
W = 123480/7.2324 J
W = 17073.2 J
W = 17.0732 kJ
W ≅ 17.07 kJ
suppose a car of 1200kg is moving with a velocity of 40km/hr therefore its kinetic energy is not zero. 1. explain briefly what happens to its kinetic energy when the driver applies the breaks and the car stops
Answer:
Explanation:
For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.
Answer:
KInetic energy reduces.
Explanation:
Application of breaks reduces velocity. Reduction of velocity constitutes velocity reduction.
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A block of mass 10kg is suspendet at a diameter of 20cm from the centre of a uniform bar im long, what force is required to balance it at its centre of gravity by applying the fore at the other end of the bar?
Answer:
4 kg of force
Explanation:
Force = (mass x distance to fulcrum) / length of fulcrum to end
Subsitute values
F = (10 x 20)/50
F =4
2.
Select the correct answer.
Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to use?
Answer:
Reverse Osmosis
Explanation:
Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.
Answer:
filtration is used to separate
why clinical thermometer cannot be used to measure the boiling point of water
Answer:
: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
George Frederick Charles Searle
Answer:
George Frederick Charles Searle FRS was a British physicist and teacher. He also raced competitively as a cyclist while at the University of Cambridge. WikipediaExplanation:
GIVE BRAINLISTTwo substances, M and N, have specific heats c and 2c. if heats Q and 4Q are supɔlied to Mand N, respectively, their changes in temperature become equal. If substance M has mass m, find the mass of substance N in terms of m
Answer:
If the mass of B is m and the temperature change is the same, the mass of B will be 2m.
Explanation:
Q = mcT
T = mc/Q
M = 4Q/2cT........... (1)
T = Q/mc
Plug this in equation 1.
M = 4Q/(2c × Q/mc) = 4Q ÷ 2Q/m = 4Q × m/2Q = 2m
Two metal spheres are made of the same material and have the same diameter, but one is solid and the other is hollow. If their temperature is increased by the same amount:_______.
A) the solid sphere becomes heavier and the hollow one lighter.
B) the solid sphere becomes bigger than the hollow one.
C) the hollow sphere becomes bigger than the solid one.
D) the two spheres remain of equal size.
E) the solid sphere becomes lighter and the hollow one heavier.
Answer:
D) the two spheres remain of equal size.
Explanation:
Since the body of the sphere is made up of both the same material. Thus the orientation will not affect the expansion. That is solid upon solid and hollow upon the hollow sphere. Hence it can be said that both the sphere expands and is due to the material used for making both of them is the same.A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
what will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between them constant
Answer:
If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
A pump lifts 400 kg of water per hour a height of 4.5 m .
Part A
What is the minimum necessary power output rating of the water pump in watts?
Express your answer using two significant figures.
Part B
What is the minimum necessary power output rating of the water pump in horsepower?
Express your answer using two significant figures.
Answer:
Power = Work / Time
P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts
Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?
Answer:
E_total = - 50 N / A
Explanation:
The electric field is a vector magnitude whereby
E_total = Eₐ + E_b
where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line
E_total = - E_a - E_b
the electric field for a point charge is
E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]
qₐ= Eₐ rₐ² / k
indicates that Eₐ = 40.0 N / C
qₐ = 40.0 0.250²/9 10⁹
qₐ = 2.777 10⁻¹⁰ C
indicates that the charge of the two points is the same
qₐ = q_b
E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²
E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]
we calculate
E_total = - 40.0 (5/4)
E_total = - 50 N / A
PLEASE HELP ME WITH THIS ONE QUESTION
A photon has 2.90 eV of energy. What is the photon’s wavelength? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
Explanation:
First, we convert the energy from eV to Joules:
[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]
[tex]= 4.64×10^{-19}\:\text{eV}[/tex]
We know from definition that
[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]
so the wavelength of the photon is
[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]
If you and a friend are standing side-by-side watching a soccer game, would you both view the motion from the same reference frame?
a. Yes, we would both view the motion from the same reference point because both of us are at rest in Earth’s frame of reference.
b. Yes, we would both view the motion from the same reference point because both of us are observing the motion from two points on the same straight line.
c. No, we would both view the motion from different reference points because motion is viewed from two different points; the reference frames are similar but not the same.
d. No, we would both view the motion from different reference points because response times may be different; so, the motion observed by both of us would be different.
Answer:
the correct is C
Explanation:
The concept of a frame of reference is of crucial importance in physics, because it is the system from which measurements are made. Therefore, the relationships between the different reference frames must be clear so that the measurements made can be compared correctly.
In this case, the first observed sees the movement of the ball, suppose it moves a distance r, the second observed is next to me, separated by a distance x, therefore a frame of reference located in the movement of the ball. ball r '.
Consequently, the measurement carried out is related by
r = r’ + x
where the bold letters indicate wind blowers.
With these explanations we review the different answers, the correct one is C
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.
Answer:
a) He found the same value of q/m for different cathode materials.
b) y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] , c) v = [tex]\frac{E_o}{B_o}[/tex]
Explanation:
In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.
b) In the part of the plates the electrons are accelerated by the electric field,
F = ma
- e E = m a
a = - (e/m) E₀
the distance traveled is
X axis
x = v₀ t
the separation of the plates is x = d
t = vo / d
Y axis
y = v_{oy} t + ½ to t²
y = ½ a t²
y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]
c) In this case there is a magnetic field B₀ and the electrons have no deflection
F = - e E + e v x B
if there is no deviation F = 0
e E = e v B
v = [tex]\frac{E_o}{B_o}[/tex]
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
