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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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u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

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According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

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But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

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Momentum before collision = Momentum after collision

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1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

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Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

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u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

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