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20 Nov 2019

Anything you can tell me about this problem that might helpismuch appreciated, thanks!!

A 2-kilogram block and an 8-kilogram block are both attachedtoan ideal spring ( for which k = 200 N/m ) are both initially atreston a horizontal frictionless surface, as shown in thediagramabove.

In an initial experiment, a 100-gram (.1 kg) ball of clayisthrown at the 2-kg block. The clay is moving horizontally wihspeedv when it hits and sticks to the block. The 8-kgblock isheld still by a removable stop. As a result, the springcompressesa maximum distance of .04 meters.

(a) Calculate the energy stored in the spring atmaximumcompression

(b) Calculate the speed of the clay ball and 2-kgblockimmediately after the clay sticks to the block but beforethespring compresses significantly.

(c) Calculate the initial speed v of the clay.

In a second experiment, an identical ball of clayisthrown at another 2-kg block, but this time the stop is removedsothat the 8-kg block is free to move.

(d) State whether the maximum compression ofthespring will be greater than, equal, or less than 0.4meters.Explain.

(e) State the principle or principles that canbeused to calculate the velcity of the 8 kg block at the instantthatthe spring regains its orginal length.

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1. A block is placed on a plane whose angle of inclination is 30˚. The coefficients of static and kinetic friction for the block on the inclined plane are both 0.2. The block
(A) remains stationary on the inclined plane.
(B) accelerates down the inclined plane.
(C) travels down the inclined plane at a constant velocity.
(D) travels up the inclined plane at a constant velocity.
(E) accelerates up the inclined plane.

2. An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30˚ with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is?

3. A 5-kg object slides down a frictionless surface inclined at an angle of 30˚ from the horizontal. The total distance moved by the object along the plane is 10 meters. The work done on the object by the normal force of the surface is?

4. A mass m = 2.5 kg is sliding along a frictionless table with initial speed v as shown in the figure. It strikes a coiled spring that has a force constant k = 500 N/m and
compresses it a distance x=5.0 cm. The initial speed v of the block was..?

5. The graph shows how the position of a particle depends on time. Which choice is closest to the instantaneous speed of the particle at t = 3s?
(A) 0.40 m/s
(B) 0.67 m/s
(C) 0.75 m/s
(D) 1.50 m/s
(E) 2.22 m/s

6. Calculate the work done by a force, F = 2x i, along the
path (0, 0) -> P₁(2, 0) -> P2(2, 2).

turquoisearmadillo973

Some Problems on Work
1. Joe is pulling a box of mass m = 10 kg up an inclined surface (?= 30degrees) using
a rope. The coefficient of kinetic friction between the surface andthe box is
Âµk = 0.5. It takes 10 seconds for Joe to bring up a box at aconstant velocity
V = 0.8m=s.
(a) What is the tension of the rope?
(b) What is the work done by the tension?
(c) What is the work done by the normal force?
(d) What is the work done by gravity?
(e) What is the total work done on the box?
(f) What is the power output necessary to Joe to bring up thebox?
maximal compression of the spring?
2. A block (m=50 kg) slides down a 3 m long inclined surface (? =30). The
coefficient of kinetic friction between the block and the surfaceis Âµk = 0.3.
Calculate the nal velocity of the block using the work-energytheorem.
3. A pendulum of length l and mass m is released from rest at anangle ? from
the equilibrium position.
(a) Neglect air friction. Calculate the velocity of the pendulumwhen the angle
is zero. Express your result in terms of ?, l, and m.
(b) Suppose ? = 60 degrees and l = 1 m. Suppose that the velocityof the pendulum
when the angle is zero is 1 m/s. Calculate the work done by the airfriction.
4. A point particle of mass m = 20 g is moving on a
at
surface when it reaches a slope which brigs it up by h = 08 m. Thespeed of
the particle in position A is v = 5 m/s.
(a) Suppose that there is no friction. What is the velocity of theparticle when
it reaches position B?
(b) Suppose that there is friction, and that the velocity of theparticle in B is
0. What is the work done by the friction from A to B?
5. A ball is shoot vertically with initial speed v, from a
table of height h.
(a) Calculate the maximal height ymax that the ball can reach (b)Calculate the velocity of the ball just before it hits thefloor.

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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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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