What is the potential energy of a 2kg plant that is on a windowsill 1.3 m high?

A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init

AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

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.

b)

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,

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

6 0

3 years ago

A metallic sphere of radius 5 cm is charged such that the potential of its surface is 100 V (relative to infinity). Draw the plo

8090 [49]

Answer:

Attached below

Explanation:

The plot showing the potential as a function of distance from the center of the sphere is attached below

note : The Potential inside the sphere will remain constant as potential remains constant on surface and it start to decrease as 1/r

6 0

2 years ago

A sound wave of frequency 440 hz travels at a speed of 344m/s through the air in a concert hall. how fast would a note one octav

Sav [38]

The speed of a sound wave does not depend on the frequency of the wave, but only on the properties of the medium.

Since we are in the same concert all, the medium of propagation is still the same (the air), so it has not changed; therefore, the speed of the sound wave has not changed, and it is still 344 m/s.

So, the correct answer is "344 m/s".

8 0

2 years ago

What is the speed of a 2000 Hz sound wave in vacuum?

Aleks [24]

A sound wave can not value in a vacuum, so its speed is 0.
Sound waves are longitudinal and require a medium to pass through.

8 0

2 years ago

Explain the force that keep the rock balanced on its tiny pedestal

Nuetrik [128]

A pedestal rock, also known as a rock pedestal or mushroom rock, is not a true balancing rock, but is a single continuous rock form with a very small base leading up to a much larger crown. Some of these formations are called balancing rocks because of their appearance. The undercut base was attributed for many years to simple wind abrasion, but is now believed to result from a combination of wind and enhanced chemical weathering at the base where moisture would be retained longest. Some pedestal rocks sitting on taller spire formations are known as hoodoos. I think this is the answer if I’m wrong I’m very sorry

5 0

2 years ago