




Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of
$2.0\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ to the right, what is the magnitude
F of the applied force?
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As shown below, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the coefficient of kinetic friction between the surface and the smaller block is 0.30. If
$F=10\phantom{\rule{0.2em}{0ex}}\text{N}$ and
$M=1.0\phantom{\rule{0.2em}{0ex}}\text{kg}$ , what is the tension in the connecting string?
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In the figure, the coefficient of kinetic friction between the surface and the blocks is
${\mu}_{\text{k}}.$ If
$M=1.0\phantom{\rule{0.2em}{0ex}}\text{kg,}$ find an expression for the magnitude of the acceleration of either block (in terms of
F ,
${\mu}_{\text{k}},$ and
g ).
$a=\frac{F}{4}{\mu}_{k}g$
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Two blocks are stacked as shown below, and rest on a frictionless surface. There is friction between the two blocks (coefficient of friction
$\mu $ ). An external force is applied to the top block at an angle
$\theta $ with the horizontal. What is the maximum force
F that can be applied for the two blocks to move together?
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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?
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A doubleincline plane is shown below. The coefficient of friction on the left surface is 0.30, and on the right surface 0.16. Calculate the acceleration of the system.
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Challenge problems
In a later chapter, you will find that the weight of a particle varies with altitude such that
$w=\frac{mg{r}_{0}{}^{2}}{{r}^{2}}$ where
${r}_{0}{}^{}$ is the radius of Earth and
r is the distance from Earth’s center. If the particle is fired vertically with velocity
${v}_{0}{}^{}$ from Earth’s surface, determine its velocity as a function of position
r . (
Hint: use
${a}^{}dr={v}^{}dv,$ the rearrangement mentioned in the text.)
$v=\sqrt{{v}_{0}{}^{2}2g{r}_{0}\left(1\frac{{r}_{0}}{r}\right)}$
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A large centrifuge, like the one shown below, is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10
g if the rider is 15.0 m from the center of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in the bottom accompanying figure. At what angle
$\theta $ below the horizontal will the cage hang when the centripetal acceleration is 10
g ? (
Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a freebody diagram of the forces to see what the angle
$\theta $ should be.)
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A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. (
Hint: since the distance traveled is of interest rather than the time,
x is the desired independent variable and not
t . Use the Chain Rule to change the variable:
$\frac{dv}{dt}=\frac{dv}{dx}\phantom{\rule{0.2em}{0ex}}\frac{dx}{dt}=v\frac{dv}{dx}.)$
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An airplane flying at 200.0 m/s makes a turn that takes 4.0 min. What bank angle is required? What is the percentage increase in the perceived weight of the passengers?
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A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance,
${F}_{\text{D}}$ , is given by the formula
${F}_{\text{D}}=\text{\u2212}bv,$ where
b is a constant and
v is the velocity. If
$b=0.750,$ and the mass of the skydiver is 82.0 kg, first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.
a. 53.9 m/s; b. 328 m; c. 4.58 m/s; d. 257 s
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In a television commercial, a small, spherical bead of mass 4.00 g is released from rest at
$t=0$ in a bottle of liquid shampoo. The terminal speed is observed to be 2.00 cm/s. Find (a) the value of the constant
b in the equation
$v=\frac{mg}{b}(1{e}^{\text{\u2212}bt\text{/}m}),$ and (b) the value of the resistive force when the bead reaches terminal speed.
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A boater and motor boat are at rest on a lake. Together, they have mass 200.0 kg. If the thrust of the motor is a constant force of 40.0 N in the direction of motion, and if the resistive force of the water is numerically equivalent to 2 times the speed
v of the boat, set up and solve the differential equation to find: (a) the velocity of the boat at time
t ; (b) the limiting velocity (the velocity after a long time has passed).
a.
$v=20.0(1{e}^{\mathrm{0.01}t});$ b.
${v}_{\text{limiting}}=20\phantom{\rule{0.2em}{0ex}}\text{m/s}$
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Questions & Answers
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
At time to = 0 the current to the DC motor is reverse, resulting in angular displacement of the motor shafts given by angle = (198rad/s)t  (24rad/s^2)t^2  (2rad/s^3)t^3
At what time is the angular velocity of the motor shaft zero
what is angular velocity
Sadiku
In three experiments, three different horizontal forces are ap
plied to the same block lying on the same countertop. The force
magnitudes are F1 " 12 N, F2 " 8 N, and F3 " 4 N. In each experi
ment, the block remains stationary in spite of the applied force.
Rank the forces according to (a) the
Sadiku
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX
state Hooke's law of elasticity
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded.
F=ke;
Shaibu
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
please help me solve this 👆👆👆
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced.
Weight of water displaced = His weight outside  his weight inside tank
= 720  34.3 = 685.7N
Now, the density of water = 997kg/m³ (this is a known value)
Volume of water displaced = Mass/Density
(next com)
Sharath
density or relative density
Shaibu
Upthrust =72034.3=685.7N
mass of water displayed = 685.7/g
vol of water displayed = 685.7/g/997
hence, density of man = 720/g / (685.7/g/997)
=1046.6 kg/m3
R.d=weight in air/upthrust in water
=720/34.3=20.99
R.d=density of substance/density of water
20.99=x/1
x=20.99g/cm^3
Shaibu
Upthrust = 72034.3=685.7N
vol of water = 685.7/g/density of water = 685.7/g/997
so density of man = 720/g /(685.7/g/997)
=1046.8 kg/m3
is there anyway i can see your calculations
Ian
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
so the density of water is 997
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x>0 but not from the left of x>0
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B)
From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope?
the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
what is the definition of model
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
matthew
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta)
therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆)
on moon, the only difference is the gravity. Gravity on moon = 0.166*g
substituting that value in R, we get the new R
Sharath
Some corrections to my old post.
Time taken to reach ground = 2*v*Sin (∆)/g
R = (2*v²*Sin(∆)*Cos(∆))/g
I put the g in the numerator by mistake in my old post. apologies for that.
R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
a ship move due north at 100kmhr1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result.
V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr.
the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
Kumar
103.077640640442km/h
Peter
state Newton's first law of motion
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia) If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Govindsingh
how would you measure displacement in your car?
The product of a. (vector b× vector a)
Source:
OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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