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3 years ago

A 3 kg rock sits on a 0.8 meter ledge. If it is pushed off, how fast will it be going at the bottom?

1 answer:

andrey2020 [161]3 years ago

6 0

As long as it sits on the shelf, its potential energy
relative to the floor is . . .

   Potential energy =    (mass) x (gravity) x (height) =

   (3 kg) x (9.8 m/s²) x (0.8m) = <u>23.52 joules</u> .

If it falls from the shelf and lands on the floor, then it has exactly that
same amount of energy when it hits the floor, only now the 23.52 joules
has changed to kinetic energy.

   Kinetic energy =    (1/2) x (mass) x (speed)²

   23.52 joules = (1/2) x (3 kg) x (speed)²

Divide each side by 1.5 kg :    23.52 m²/s² = speed²

Take the square root of each side: speed = √(23.52 m²/s²) = <em>4.85 m/s </em> (rounded)

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

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

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

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

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

3 years ago

Read 2 more answers

17. _______ is a fuel produced by fermenting crops.

DENIUS [597]

Fuel is produced by fermenting Ethanol

6 0

3 years ago

Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different

kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

Solution:

- The standard atmospheric equation is expressed as:

P = P_a* ( 1 - βh/T_a)^(g / R*β)

(g / R*β) = 32.174 / 1716*0.0035 = 5.252

P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

P = 1240 lb/ft^2

- The air density method which is expressed as:

P = P_a - γ*h

P = 2116.2 - 0.07647*14,110

P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

P = P_a* e^ ( -g*h / R*T_a )

P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

P = 1270 lb/ft^2

6 0

2 years ago

Read 2 more answers

Which best describes what occurs when an object takes in a wave as the wave hits it?.

statuscvo [17]

Answer:

Absorption

Explanation:

7 0

2 years ago

What is a derived physical quantity? Name three derived physical quantities, and for each, give its S.I. units and its U.S. Cust

anastassius [24]

Answer:

Physical quantity is a physical property of an object or material that can be expressed by magnitude and unit.

The derived physical quantities are the type of physical quantities which can be expressed or defined by other physical quantities, called the base quantities. Example: Area, Volume, Velocity

Area- SI Unit: m², U.S. Customary unit: acre

Volume- SI Unit: m³, U.S. Customary unit: cubic inch

Velocity- SI Unit: m/s, U.S. Customary unit: ft/s

6 0

3 years ago