Physics, 31.01.2020 16:53 zarek93

Acompact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible friction. the object is pulled to a distance of 0.200 m from its equilibrium position, held in place with a force of 16.0 n, and then released from rest. it then oscillates in simple harmonic motion. (the object oscillates along the x-axis, where x = 0 is the equilibrium position.) (a) what is the spring constant (in n/m)? n/m (b) what is the frequency of the oscillations (in hz)? hz (c) what is the maximum speed of the object (in m/s)? m/s (d) at what position(s) (in m) on the x-axis does the maximum speed occur? x = ± m (e) what is the maximum acceleration of the object? (enter the magnitude in m/s2.) m/s2 (f) at what position(s) (in m) on the x-axis does the maximum acceleration occur? x = ± m (g) what is the total mechanical energy of the oscillating spring–object system (in j)? j (h) what is the speed of the object (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s (i) what is the magnitude of the acceleration of the object (in m/s2) when its position is equal to one-third of the maximum displacement from equilibrium? m/s2

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Answer from: daniel1480

(a) 80 N/m

The spring constant can be found by using Hooke's law:

F=kx

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

$k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m$

(b) 0.84 Hz

The frequency of oscillation of the system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz$

The maximum speed of a spring-mass system is given by

$v=\omega A$

where

$\omega$ is the angular frequency

A is the amplitude of the motion

For this system, we have

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

A=0.200 m (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

v=(5.25 rad/s)(0.200 m)=1.05 m/s

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

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

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

$a=\omega^2 A$

Using the numbers we calculated in part c):

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

A=0.200 m

we find immediately the maximum acceleration:

a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2

(f) At the position of maximum displacement: $x=\pm 0.200 m$

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

$a=\frac{F}{m}$

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

F=kx

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

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

where

m = 2.90 kg is the mass

$v_{max}=1.05 m/s$ is the maximum speed

Solving for E, we find

$E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J$

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

so the elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J$

and since the total energy E = 1.60 J is conserved, the kinetic energy is

K=E-U=1.60 J-0.18 J=1.42 J

And from the relationship between kinetic energy and speed, we can find the speed of the system:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s$

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

So the restoring force exerted by the spring on the mass is

F=kx=(80 N/m)(0.0667 m)=5.34 N

And so, we can calculate the acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2$

Answer from: Quest

La olla es de metal por que se calienta mas rápido, y las manijas son de madera por que así no se calientas y tienden a estar mas frías.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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