PHY 2048 Midterm: 2048_Fall13_Exam2_WithSolutions

18 views4 pages

turquoisegnu128

31 Jan 2019

School

University of Florida

Department

Physics

Course

PHY 2048

Professor

All

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

2048_fall13_exam2: solutions to 2nd exam: a particle starts from rest at time t = 0 and moves along the x axis. Meanwhile, the d mk d mk to say jt f jt dt jt dt solve for p mk mk mk mv. 4 t (1. 2: an object attached to a cord is moving in a circular path of radius 0. 5m on a horizontal frictionless surface. The cord will break if its tension exceeds 16 n. the maximum kinetic energy the object can have is: (ans = 4j). 2 r r r r f r r. The work done by the external force as the mass makes one revolution is (ans = 47 j). Always tangent to the track means the angle between the force and the displacement is 0 degrees, 47 n m : the potential energy of a body of mass m is given by u = mgx + kx2/2.

Related Questions

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

1
2
3
4
5
6
7
8

puceworm982

Good Vibes: Introduction to Oscillations

The conditions that lead to simple harmonic motion are as follows:

There must be a position of stable equilibrium.

There must be a restoring force acting on the oscillating object. The direction of this force must always point toward the equilibrium, and its magnitude must be directly proportional to the magnitude of the object's displacement from its equilibrium position. Mathematically, the restoring force F⃗ is given by ⃗, where is the displacement from equilibrium and k is a constant that depends on the properties of the oscillating system.

The resistive forces in the system must be reasonably small.

Consider a block of mass m attached to a spring with force constant k, as shown in the figure(Figure 1) . The spring can be either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When the spring is relaxed, the block is located at x=0. If the block is pulled to the right a distance A and then released, A will be the amplitude of the resulting oscillations.

Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block.

Part A

After the block is released from x=A, it will

(a) remain at rest

(b) move to the left until it reaches equilibrium and stop there.

(d) move to the left until it reaches x=−A and then begin to move to the right.

The time it takes the block to complete one cycle is called the period. Usually, the period is denoted T and is measured in seconds.

The frequency, denoted f, is the number of cycles that are completed per unit of time: f=1/T. In SI units, f is measured in inverse seconds, or hertz (Hz).

Part B

If the period is doubled, the frequency is

(a) unchanged

(b) doubled

Part C

An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. What is its frequency f?

Express your answer in hertz.

Part D

If the frequency is 40 Hz, what is the period T?

Express your answer in seconds.

The following questions refer to the figure (Figure 2) that graphically depicts the oscillations of the block on the spring.

Note that the vertical axis represents the x-coordinate of the oscillating object, and the horizontal axis represents time.

Part E

Which points on the x-axis are located at a distance A from the equilibrium position?

(a) R only

(b) Q only

Part F

Suppose that the period is T. Which of the following points on the t-axis are separated by the time interval T?

(a) K and L

(b) K and M

(d) L and N

(e) M and P

Now assume for the remaining Parts G - J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s.

Part G

What is the period T?

Express your answer in seconds.

Part H

How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement?

Express your answer in seconds.

Part I

What distance d does the object cover during one period of oscillation?

Express your answer in meters.

Part J

What distance d does the object cover between the moments labeled K and N on the graph?

Express your answer in meters.

emeraldyak680

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.

2 Answers

Relevance

electron1
Lv 7

4 years ago

There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’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. Let’s use the following equation to determine the acceleration.

vf2 = vi^2 + 2 * a * d

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

24.8 * a = 40.29

a = 40.29 ÷ 24.8

This is approximately 1.62 m/s^2.

Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

This is approximately 82 N.

Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

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

Net force = F + (1187.76 ÷ 12.4) – 41

Set the two net forces equal to each other.

F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

F is approximately 27.3 N

Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J
JupiterSky
Lv 6

4 years ago

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

while the friction does negative work.

∆K = ∆U +∆W + ∆Wf

½m(Vf²−Vi²) = mgh + ∆W + Ff*d

∆W = ½m(Vf²−Vi²) − mgh − Ff*d

∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

∆W = 508 J

Still have questions? Get your answers by asking now.

Next Question

---

asked in Science & MathematicsPhysics · 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 ﬁgure below). The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track
(as shown in the ﬁgure below). It leaves the
track horizontally, ﬂies through the air, and
subsequently strikes the ground.
acceleration = 9.81
 
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

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

3 Answers

Relevance

Jánošík
Lv 7

8 years ago

Favorite Answer

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

Edit:

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

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

t = √( 2h / g )

t = √[ 2×(1.9 m) / (9.81 m/s²) ]

t = 0.6224 s

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

r = v×t

(4.59 m) = v×(0.6224 s)

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

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

Its kinetic energy at that time is:

KE = m×v²/2

KE = (0.487 kg)×(7.3749 m/s)²/2

KE = 13.244 J

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

KE = PEe = k×d²/2

(13.244 J) = k×(0.047 m)²/2

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

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

PEg = m×g×h

PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

PEg = 9.077 J

The total energy of the block on the table is:

TE = PEg + PEe

TE = (9.077 J) + (13.244 J)

TE = 22.32 J

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

TE = KE = m×v²/2

(22.32 J) = (0.487 kg)×v²/2

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

...Show more
Technobuff
Lv 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".??
STALKER
Lv 4

8 years ago

use f=kl

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

k=f/l

since you have mass in grams. use w=mg

w=0.487 x 9.8 = 4.77N

Now use the equation,

k=4.77/4.7

k=1.012

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

Still have questions? Get your answers by asking now.

Next Question

ひいらぎ

Lv 5

asked in Science & MathematicsPhysics · 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.
 
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

Physics Momentum, Rolling?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

I can't visualize the first problem.

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.

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

Still have questions? Get your answers by asking now.

Next Question

asked in Science & MathematicsPhysics · 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.
 
(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
 
(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

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

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

♣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;

♠ kinetic energy of the pig at the moment it’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;

♦pig’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=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

f= μ*mg*cos(b);

♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

(μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

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

♥ pig’s potential energy above the low end of the ramp is

W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

Pig’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) – μ*mg*cos(b)*(s + h/sin(b));

v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

(b); v =1.605 m/s;

...Show more

Still have questions? Get your answers by asking now.

+100

Join Yahoo Answers and get 100 points today.

PHY 2048 Midterm: 2048_Fall13_Exam2_WithSolutions

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHY 2048 Midterm: 2048_Spring17_Exam2_Solution

PHY 2048 Midterm: 2048_Fall16_Exam1

PHY 2048 Midterm: 2048_Spring17_Exam1

Related Questions

2 Answers

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

3 Answers

Physics Momentum, Rolling?

1 Answer

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

1 Answer

Trending Questions