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23 Nov 2019

Zero Net Force A spaceship is on a straight-line path between Earthand its moon. At what distance from Earth is the net gravitationalforce (due to the Earth and Moon only) on the spaceship zero?
(answer)km

Section 14.3 Gravitation and Superposition

I figure that I would have to use Earth's mass = 5.9E+24kg, andGravitational Force = 6.67E-11Nm^2/kg^2
Fnet = (Gm1*m2)/r^2

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In the example 12.5 (Section 12.3) in the textbook we ignored thegravitational effects of the moon on a spacecraft en route from theearth to the moon. In fact, we must include the gravitationalpotential energy due to the moon as well. For this problem, you canignore the motion of the earth and moon.

Part A
If the moon of mass m_{\rm M} has radius R_{\rm M} and the distancebetween the centers of the earth and the moon is R_{\rm EM}, findthe total gravitational potential energy of the particle-earth andparticle-moon systems when a particle with mass m is between theearth and the moon, and a distance r from the center of the earth.Take the gravitational potential energy to be zero when the objectsare far from each other. Take the mass of earth as m_{\rm E}.

Part B
There is a point along a line between the earth and the moon wherethe net gravitational force is zero. Use the expression derived inpart (a) to find the distance of this point from the center of theearth.

Part C
With what speed must a spacecraft be launched from the surface ofthe earth just barely to reach this point?

Part D
If a spacecraft were launched from the earth's surface toward themoon with an initial speed of 11.2 {\rm km/s}, with what speedwould it impact the moon?

(a) What is the weight of a 5.78 kg backpack at the surface ofthe Earth?
56.7 N ( I understand this problem after some time :) )
(b) What is the weight of the same backpack at the surface of theMoon where the acceleration due to gravity is only 1.6 m/s2?
9.37 N (as well as this problem)

(c) If the backpack is atthe surface of the Earth, and if a net force of 8 N is applied tothe backpack, what is the amount of acceleration of thebackpack?
(Not sure here)

m/s^2
I was thinking A = Fnet/m
Fnet being the total force on the object being equal to G(mass ofthe earth)/(radius of the earth)^2
(I know it isn't as simple as adding 8N to the weight of backpackat the surface of the earth, if I divide by the net force thatisn't adding the net force to the weight...Please Help)

(d) If the backpack is atthe surface of the Moon, and if the same net force is applied tothe backpack, what is the amount of acceleration of thebackpack?
(If I don't understand C of course I am confused on this problem aswell, thank you for your help in advance I can figure out where togo with this problem if I understand C)

m/s2
Please explain to me so that I understand for test

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

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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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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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