Physics, 24.03.2020 19:41 maustin5323

A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an initial compression of the spring of 7 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion?
(b) What is the maximum speed during the motion?

Answers: 1

Show answers

Answers

Answer from: fonsworth5

a) 19.4 cm

b) 3.2 m/s

Explanation:

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

So the total mechanical energy of the system is

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

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

$E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J$

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

$E=U=\frac{1}{2}kx_{max}^2$

where $x_{max}$ is the maximum stretch. Solving for $x_{max}$ ,

$x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m$

So, 19.4 cm.

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$

where

m is the mass

$v_{max}$ is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

$v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s$

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

Answer from: Quest

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

Another question on Physics

Physics, 21.06.2019 23:30

Two technicians are discussing a resistance measurement between the can-h and can-l wires. technician a says this measurement should be done with the ignition switch in the "run" position. technician b states that a measurement of 0 ohms indicates an open in the network. which technician is correct?

Answers: 3

Answer

Physics, 22.06.2019 16:00

What part of the ear is names after tools, such as the hammer and the anvil?

Answers: 1

Answer

Physics, 22.06.2019 19:30

Ashot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. the shot hits the ground 2.08 s later. you can ignore air resistance. how far did she throw the shot?

Answers: 2

Answer

Physics, 22.06.2019 22:30

Find v1, the speed of the fluid in the left end of the main pipe. express your answer in terms of h1, h2, g, and either a1 and a2 or γ, which is equal to a1a2.

Answers: 2

Answer

You know the right answer?

A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The sys...

Questions