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10 Nov 2019

I need follow up help with the following question:

Here was the complete response from the guy who helped me:

Here is my question. I don't understand how they got theequation for Fx and Fy.

Fx=k[(-1/.2)-(cos(45)/.283)]

Fy=k[(1/.2)-(cos(45)/.283)]

Is the .2 from the distance 2cm? And, how did he get .283?

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Hubert KochLv2

17 Oct 2019

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A mass m rests on a frictionless horizontal table and isconnected to rigid supports via 2 identical springs each of relaxedlength L0 and spring constantk. Each spring is stretched to a length L considerable greater thanL0. Horizontal displacementsof m from its equilibrium position are labeled x and y.

(Diagram insert didn't work, all it really tells you is that xis on the axis of the springs and y is perpendicular to thesprings)

a) Write down the differential equation of motion (i.e.,Newton's Law) governing small oscillations in the xdirection.

b) Same as a but in the y direction (assume that y <<L).

c) In terms of L and L0, calculate the ratio of the periods ofthe oscillations along x and y.

d) If at t=0 the mass m is released from the point x = y = A0with zero velocity, what are its x and y coordinates at any latertime t?

---------------------------

So, I started with just ma = -2kx for a, leading me to thedifferential equation x''(t) + (2k/m) x = 0.

Then I did basically the same thing with y.

However, when I get to c, there's my problem - I don't knowhow I could write any of this in L and L0.

Also, b doesn't really make sense to me because if y is smallwhen compared to L, it seems like there shouldn't be anyoscillation in the y direction.

The answer the back of the book shows for c and d are:

c) Tx/Ty = (1 - L0 / L)1/2

d) x(t) = A0cos( (2k/m)1/2 t)

y(t) = A0cos( (2k(L-L0)/mL)1/2 t)

I have no idea how they got to these answers. Can somebodyhelp me? It has something to do with springs already beingstretched, but it seems they are making an assumption somewherethat I don't understand.

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.

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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

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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

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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

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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

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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

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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

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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

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A wheel 75 meters in diameter is mounted vertically on an infinitefrictionless plane at the origin. The infinite frictionless planehas a grid of Cartesian coordinates on it. The moment of inertia ofthe wheel is mr^2. The mass of the wheel is 2000kg. There is anengine connected to the wheel that can produce a torque of 2.34x10^6 Nm. There is a 50 kg object attached to the circumference ofthe wheel that is attached to the wheel with a remote releasemechanism. The wheel is placed on the origin so that the plane ofthe wheel contains the point x=50 meters and y=86.8 m. The wheel ispositioned so that the object is just above the plane at the 6o'clock position. The engine is engaged and begins to apply thetorque to the wheel at 12:00 noon on March 1, 2011. The wheelbegins to rotate in the counterclockwise direction. The torque isapplied for 4 x 10^4 revolutions. At that point the object isreleased from the wheel and a 75 Newton motor accelerates theobject in the direction it is released. The motor is fired for 2.34x 10^4 seconds. (the following questions i have already answered)What is the angular velocity and rotational KE of the object justbefore it is released? (answer for that is 6.39 x 10^2 rad/sec and5.89 x 10^11 J). What angle does the path of the object make withthe x axis? (answer is 60.05 degrees). Where is the object on theinfinite frictionless plane when the motor stops (what are itscoordinates....answer is x=4.85x10^8m y=8.40x10^8m). What is thevelocity and linear KE of the object when the motor stops?(5.91x10^4 m/sec and KE= 8.73 x 10^10J). How much work was done onthe object up until the time the motor stopped? (8.73 x 10^10J).(Hint: the mass of the wheel is 2050 kg while the torque is appliedand the object is attached).

YEA I know what you are thinking at this moment. Difficult right?Please give it a chance and have faith in yourself. Here is thepart I had difficulty with:

After the engine stops firing the object is turned around so thatthe engine can be used to stop the object. The motor is fired atits rated thrust to stop the object. Unfortunately it malfunctionsand shuts down after an hour and the object coasts from then on. Atthe end of an hour of coasting it encounters a ramp that is 2000meters high and slopes up at an angle of 20 degrees. The objectclimbs the ramp and flies off the ramp into the air when it reachesthe end of the ramp. It strikes the plane and digs a crater 5meters deep in the plane and comes to a stop. How far from the endof the ramp does it land? Where does it land on the plane (as incoordinates)? What is the resultant velocity of the object and whatis the acute angle of the resultant velocity with the plane when itstrikes the plane? What is the KE of the object when it hits theplane? What date and time is it when the object lands on the plane?How much work was done on the object to stop it? How long did ittake to stop? What was the deceleration of the object in m/sec^2?What was the force on the object in Newtons to stop it?

If you are from MIT I have faith that you can do this. I am only ajunior in high school taking first year physics that is not evenAP. My teacher is from MIT and said he did this kind of stuff whenhe was there. Please please please help me! I'm desperate.

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