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Rasek [7]
3 years ago
13

When is the velocity of a mass on a spring at its maximum value?

Physics
2 answers:
ehidna [41]3 years ago
8 0

Answer:

A.  when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.


OLEGan [10]3 years ago
6 0

<em>Option (A) is correct. The spring-mass system has the maximum velocity</em> <em><u>when the displacement of the mass is zero. </u></em>

Further Explanation:

A spring mass system oscillating with small amplitude is considered to be executing simple harmonic motion. The energy of he spring mass system during its oscillation remains conserved.

When the spring is completely compressed, the whole energy of the system is stored in the form of spring potential energy and as the spring starts to come to its natural length, the spring potential energy starts to convert to the kinetic energy of the mass.

When the spring crosses its natural length and stretches further, again the mass losses its energy and spring potential energy is stored until the mass stops and complete kinetic energy is converted into spring potential energy.

At any point, the energy of the spring mass system can be expressed as:

\begin{aligned}{E_{total}}&= KE + PE\\&=\frac{1}{2}m{v^2} + \frac{1}{2}k{x^2}\\\end{aligned}  

Now, in order to obtain the maximum velocity during its motion, the mass should have the whole of the energy of the system I the form of kinetic energy. To maximize the kinetic energy, the spring potential energy should be zero.

The spring potential energy of the system will be zero at the natural length of the spring or at the point where the displacement of the mass is zero from the natural lengthy of the spring.

Thus,<em> </em><em>option (A) is correct.</em><em> </em><em>The spring-mass system has the maximum velocity</em> <em><u>when the displacement of the mass is zero. </u></em>

<em><u> </u></em>

Learn More:

1. You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track brainly.com/question/10177389

2. It's been a great day of new, frictionless snow. Julie starts at the top of the 60∘ slope shown brainly.com/question/3943029

3. The amount of kinetic energy an object has depends on its brainly.com/question/137098

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Spring-mass system

Keywords:

Spring-mass system, spring potential energy, kinetic energy of mass, total energy, natural length, compressed, maximum speed, displacement zero, maximum kinetic energy.

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r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

KE=\frac{1}{2}mv^2

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

KE=\frac{1}{2}mv^2

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

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4 0
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Answer:

<h2>17.1 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

3800 g = 3.8 kg

We have

force = 3.8 × 4.5

We have the final answer as

<h3>17.1 N</h3>

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3 years ago
11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
Marat540 [252]

a =  \frac{v - u}{t}

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

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the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

5.0 m/s^2

5.0 > 3.33 \\ so \:  the \: answer  \: is \: no

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Answer:

No

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The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

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The maximum speed the mass can have before it breaks is calculated as follows;

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Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

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