1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tino4ka555 [31]
3 years ago
6

A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a sp

eed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m

Physics
1 answer:
cluponka [151]3 years ago
7 0

Answer:

(a) Approximately 0.335\; \rm m.

(b) Approximately 1.86\; \rm m\cdot s^{-1}.

(c) Approximately 0.707\; \rm m.

(d) Approximately 0.228\; \rm m.

Explanation:

  • v_i denotes the velocity of the object in the first diagram right before it came into contact with the spring.
  • Let m denote the mass of the block.
  • Let \mu denote the constant of kinetic friction between the object and the surface.
  • Let g denote the constant of gravitational acceleration.
  • Let k denote the spring constant of this spring.
<h3>(a)</h3>

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy \displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2.

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of D and compressed the spring by the same distance.

  • Energy lost to friction: \underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D.
  • Elastic potential energy that the spring has gained: \displaystyle \frac{1}{2}\,k\, D^2.

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2.

Assume that g = 9.81\; \rm m \cdot s^{-2}. In the equation above, all symbols other than D have known values:

  • m =1.10\; \rm kg.
  • v_i = 2.60\; \rm m \cdot s^{-1}.
  • \mu = 0.250.
  • g = 9.81\; \rm m \cdot s^{-2}.
  • k = 50.0\; \rm N \cdot m^{-1}.

Substitute in the known values to obtain an equation for D (where the unit of D\! is m.)

3.178 = 2.69775\, D + 25\, D^2.

2.69775\, D + 25\, D^2 + 3.178 = 0.

Simplify and solve for D. Note that D > 0 because the energy lost to friction should be greater than zero.

D \approx 0.335\; \rm m.

<h3>(b)</h3>

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J.

As the object moves to the left, part of that energy will be lost to friction:

(\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J.

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J.

Calculate the velocity corresponding to that kinetic energy:

\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}.

<h3>(c)</h3>

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy (1.91\; \rm J) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is \mu \cdot m \cdot g.

\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m.

<h3>(d)</h3>

Similar to (a), solving (d) involves another quadratic equation about D.

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) 1.91\; \rm J.

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2.

25\, D^2 + 2.69775\, D - 1.90811\approx 0.

Again, D > 0 because the energy lost to friction is greater than zero.

D \approx 0.228\; \rm m.

You might be interested in
Which of the following statements is TRUE about updating the exposure control plan?
iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

<em>Updates must have the  reflection of changes in tasks as well in procedures.</em>

<em>Updates must reflect changes in positions that affect occupational exposure.</em>

<em>Updates must have the cost of PPE that is needed and  necessary to reduce exposure</em>

An exposure control plan can be regarded as  the framework for compliance between the employer and the workers.

  • This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

  • This plan gives hope to workers in term of protection when working with their Employer.

  • There are some elements that is associated with  Exposure Control Plan, and theses are;
  1. Health hazards as well as  risk that is attributed to  each product in the worksite.
  2. Statement of purpose.
  3. procedures and practices in a written form
  4. Responsibilities from the Manager, CEO, designated resources and employer.

Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.

brainly.com/question/1203927?referrer=searchResults

3 0
2 years ago
An object rolls East at a steady speed of 12 m/s for 3 seconds. What distance did the object travel
horrorfan [7]
Answer: 36 meters.


Equation to find distance:
Speed x time
8 0
3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
How many earth years does it take mars to orbit the sun
Colt1911 [192]

This is a question that would have literally have taken two seconds to look up on google but the answer is 1.88 years.

4 0
2 years ago
Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut
Harman [31]

Answer: a) 11.76 m/s  b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

4 0
3 years ago
Other questions:
  • Have scientists discovered a fifth fundamental force of nature?
    8·1 answer
  • According to the conservation of momentum, if a small football player collides with a much larger football player, which of the
    15·1 answer
  • car accelerates from 10m/s to 25m/s in 6.0s what is its acceleration? how far did it travel in this time? assume constant accele
    6·1 answer
  • A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re
    10·1 answer
  • You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
    6·1 answer
  • Why is the specific heat of salt solution lower than that of water?
    10·1 answer
  • Explain the necessity of smoothing the output voltage before applying it to a transistor
    13·1 answer
  • DontBlink1196 is *HEDJL:::A:APSKDL
    11·1 answer
  • What is food called when it enters the pharynx?<br><br> A.Bile <br> B.bolus<br> C. Food<br> D. Feces
    5·1 answer
  • a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!