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PHYS114 Study Guide - Midterm Guide: Friction, European Route E6, Terminal Velocity

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turquoisegnu128
24 Oct 2018
School
University of Alberta
Department
Physics
Course
PHYS114
Professor
Al Meldrum

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Document Summary

Time: 50 minutes; calculators & formulae provided at the end = only aid; total = 20 points. 1 a)[3] the coe cient of static friction between a crate and the (level) oor is given as s = 0. 4, while the kinetic friction coe cient is s = 0. 15. The mass of the crate is m = 55kg. You apply a force that is just enough to make the refrigerator move. 1 b)[2] the maximum speed at which you will push this crate is vf = 5. 0m/s. Provide a velocity vs time graph from the start, and past the point of reaching vf. 2 a) [3] a girl (mass m) is jumping from a ladder. Indicate the forces with arrows of appropriate respective length in the fb diagrams, and also the acceleration of the centre of gravity (beside the diagrams). 7. 8 103, glass = 2. 5 103, lead = 1. 1 104.

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Physics: Principles with Applications
7th Edition, 2013
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ISBN: 9780321625922

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Related Questions

A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.



ANGULAR MOMENTUM
(a) Consider the system consisting of the child and the disk, butnot including the axle. Which of the following statements are true,from just before to just after the collision?
The momentum of the system doesn't change.
The momentum of the system changes.
The axle exerts a force on the system but nearly zero torque.
The torque exerted by the axle is zero because the force exerted bythe axle is very small.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardlychanges.
The torque exerted by the axle is nearly zero even though the forceis large, because || is nearly zero.



(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kgƂĀ·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kgƂĀ·m2/s
(d) If the disk was initially at rest, now how fast is it rotating?That is, what is its angular speed? (The moment of inertia of auniform disk is ƂĀ½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.



MOMENTUM
(g) What was the speed of the child just after the collision?
v = m/s
(h) What was the speed of the center of mass of the disk just afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kgƂĀ·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kgƂĀ·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.



ANGULAR MOMENTUM
(a) Consider the system consisting of the child and the disk, butnot including the axle. Which of the following statements are true,from just before to just after the collision?
The momentum of the system doesn't change.
The momentum of the system changes.
The axle exerts a force on the system but nearly zero torque.
The torque exerted by the axle is zero because the force exerted bythe axle is very small.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardlychanges.
The torque exerted by the axle is nearly zero even though the forceis large, because || is nearly zero.



(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kgƂĀ·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kgƂĀ·m2/s
(d) If the disk was initially at rest, now how fast is it rotating?That is, what is its angular speed? (The moment of inertia of auniform disk is ƂĀ½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.



MOMENTUM
(g) What was the speed of the child just after the collision?
v = m/s
(h) What was the speed of the center of mass of the disk just afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kgƂĀ·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kgƂĀ·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

Ā 

Ā 
Ā 
Ā 

2 Answers

Relevance
  • electron1
    LvĀ 7
    4 years ago

    There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchairā€™s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Letā€™s use the following equation to determine the acceleration.

    Ā 

    vf2 = vi^2 + 2 * a * d

    6.5^2 = 1.4^2 + 2 * a * 12.4

    24.8 * a = 40.29

    a = 40.29 Ć· 24.8

    This is approximately 1.62 m/s^2.

    Ā 

    Net force = 50.5 * 40.29 Ć· 24.8 = 2034.645 Ć· 24.8

    This is approximately 82 N.

    Ā 

    Force parallel = 50.5 * 9.8 * 2.40 Ć· 12.4 = 1187.76 Ć· 12.4

    Ā 

    This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

    Ā 

    Net force = F + (1187.76 Ć· 12.4) ā€“ 41

    Set the two net forces equal to each other.

    Ā 

    F + (1187.76 Ć· 12.4) ā€“ 41 = 2034.645 Ć· 24.8

    F = 41 + (2034.645 Ć· 24.8) ā€“ (1187.76 Ć· 12.4)

    F is approximately 27.3 N

    Ā 

    Work = [41 + (2034.645 Ć· 24.8) ā€“ (1187.76 Ć· 12.4)] * 12.4 = 337.9625 J

    Ā 
  • JupiterSky
    LvĀ 6
    4 years ago

    His change in kinetic energy is due to gravitational- and his own work,

    while the friction does negative work.

    Ā 

    āˆ†K = āˆ†U +āˆ†W + āˆ†Wf

    Ā½m(VfĀ²āˆ’ViĀ²) = mgh + āˆ†W + Ff*d

    āˆ†W = Ā½m(VfĀ²āˆ’ViĀ²) āˆ’ mgh āˆ’ Ff*d

    āˆ†W = Ā½*50.5(6.5Ā²āˆ’1.4Ā²) āˆ’ (50.5*9.8*2.4) āˆ’ (āˆ’41*12.4)

    āˆ†W = 508 J

    Ā 
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Ā 
Next Question
---
asked inĀ Science & MathematicsPhysicsĀ Ā·Ā 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the

spring (located on the left-hand side of the

track) and compresses the spring a distance

4.7 cm from its equilibrium position (as shown

in the ļ¬gure below). The block starts from

rest, is accelerated by the compressed spring,

and slides across a frictionless horizontal track

(as shown in the ļ¬gure below). It leaves the

track horizontally, ļ¬‚ies through the air, and

subsequently strikes the ground.

acceleration = 9.81

Ā 

A) What is the spring constant?

B)What is the speed v of the block when it leaves

the track?

C) What is the total speed of the block when it

hits the ground?

...Show more
Update:

Height Above ground = 1.9 m

Lands = 4.59m away

Ā 
What is the spring constant? please help me!!!!?

3 Answers

Relevance
  • JĆ”noÅ”Ć­k
    LvĀ 7
    8 years ago
    Favorite Answer

    I'm sure there is some crucial information in "the figure below" !!!

    Ā 

    Edit:

    OK... now we got something to work with ... ;-)

    Ā 

    Time needed to fall a vertical distance of 1.9 m is:

    t = āˆš( 2h / g )

    t = āˆš[ 2Ɨ(1.9 m) / (9.81 m/sĀ²) ]

    t = 0.6224 s

    Ā 

    In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

    r = vƗt

    (4.59 m) = vƗ(0.6224 s)

    v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

    Ā 

    That is also the speed of the block as it leaves the spring.

    Its kinetic energy at that time is:

    KE = mƗvĀ²/2

    KE = (0.487 kg)Ɨ(7.3749 m/s)Ā²/2

    KE = 13.244 J

    Ā 

    That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

    KE = PEe = kƗdĀ²/2

    (13.244 J) = kƗ(0.047 m)Ā²/2

    k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

    Ā 

    The gravitational potential energy of the block on the table is:

    PEg = mƗgƗh

    PEg = (0.487 kg)Ɨ(9.81 m/sĀ²)Ɨ(1.9)

    PEg = 9.077 J

    Ā 

    The total energy of the block on the table is:

    TE = PEg + PEe

    TE = (9.077 J) + (13.244 J)

    TE = 22.32 J

    Ā 

    When hitting the ground, all that energy is converted to kinetic energy, so

    TE = KE = mƗvĀ²/2

    (22.32 J) = (0.487 kg)ƗvĀ²/2

    v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

    ...Show more
    Ā 
  • Technobuff
    LvĀ 7
    8 years ago

    There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??

    Ā 
  • STALKER
    LvĀ 4
    8 years ago

    use f=kl

    Ā 

    where k is the spring constant and l is extension from the natural length.

    Ā 

    k=f/l

    Ā 

    since you have mass in grams. use w=mg

    Ā 

    w=0.487 x 9.8 = 4.77N

    Ā 

    Now use the equation,

    Ā 

    k=4.77/4.7

    Ā 

    k=1.012

    Ā 

    Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

    ...Show more
    Ā 
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Next Question
ć²ć„ć‚‰ćŽ
LvĀ 5
asked inĀ Science & MathematicsPhysicsĀ Ā·Ā 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.

Ā 

A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.

-What is the moment of Inertia of the wash load?

-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2

...Show more
Ā 
Physics Momentum, Rolling?

1 Answer

Relevance
  • Ā 
    Anonymous
    1 decade ago
    Favorite Answer
    Ā 

    I can't visualize the first problem.

    Ā 

    The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

    Ā 

    Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

    ...Show more
    Ā 
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Next Question
AZ
asked inĀ Science & MathematicsPhysicsĀ Ā·Ā 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.

Ā 

(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree

Ā 

(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s

...Show more
Ā 
Physics:You and your friends push a 75-kg...Help!?

1 Answer

Relevance
  • Ā 
    Anonymous
    1 decade ago
    Favorite Answer
    Ā 

    ā™£the distance the pig was pushed up is

    s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

    speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

    ā™  kinetic energy of the pig at the moment itā€™s released after push-up is

    E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

    ā™¦pigā€™s potential energy at vertex (when stopped) is E1=mgh;

    the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

    f=Ī¼*w1, w1=mg*cos(b) is component of pigā€™s weight normal to the aluminum ramp; or;

    f= Ī¼*mg*cos(b);

    ā™¦Thus E2=E-E1; or; f*L = 0.5*m*v^2 ā€“mgh; or;

    (Ī¼*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as ā€“mgh; or;

    Ī¼*gh*cos(b)/sin(b) =as ā€“gh, hence cot(b) =(as ā€“gh)/(Ī¼*gh), hence

    (a); b=atan(Ī¼*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5Ā°;

    Ā 

    ā™„ pigā€™s potential energy above the low end of the ramp is

    W= mg*(s*sin(b) +h); pigā€™s ramp position is p=s +L = s + h/sin(b);

    Pigā€™s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

    and W1=0.5*m*v^2; thus

    0.5*m*v^2 = mg*(s*sin(b) +h) ā€“ Ī¼*mg*cos(b)*(s + h/sin(b));

    v^2 =2g*(s*sin(b) +h)*(1 ā€“Ī¼*cot(b)) =2.575;

    (b); v =1.605 m/s;

    ...Show more
    Ā 
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