What force pulled the leaf to the ground

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

Which takes more energy, going from 0 to 30, or from 30 to 60?

AfilCa [17]

0 to 30 takes more energy

3 0

3 years ago

. How much power is used to move a 5 kg mass 28 meters in 8 seconds?

saul85 [17]

Answer:

171.5Watts

Explanation:

Given parameters:

Mass = 5kg

Distance = 28m

Time = 8s

Unknown:

Power = ?

Solution:

Power is the rate at which work is done;

Power = $\frac{work done }{time taken}$

Work done = force x distance = mg x d =

m is the mass

g is the acceleration due to gravity

d is the distance

Work done = 5 x 9.8 x 28 = 1372J

Power = $\frac{1372}{8}$ = 171.5Watts

6 0

3 years ago

Wave energy can only be transmitted through a material medium. wave energy can only be transmitted through a material medium.

S_A_V [24]

The correct answer is False.
In fact, electromagnetic waves (which carry energy), do not need a material medium to travel, because they can travel in vacuum as well. So, wave energy can be transmitted also through vacuum.

3 0

3 years ago

The magnetic flux through a metal ring varies with time t according to ΦB = at3 − bt2, where ΦB is in webers, a = 6.00 Wb s−3, b

pychu [463]

To solve this problem it is necessary to apply the second derivative of the function to find the maximum time reference and thus calculate the maximum voltage.

With the maximum voltage by Ohm's Law it is possible to find the maximum current.

Ohm's law defines that

E = I*R

Where,

I = Current

R= Resistance

On the other hand by faraday studies and the potential can be expressed at the rate of change of the electric flow, that is

$E = \frac{d\phi}{dt}$