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Lilit [14]
3 years ago
13

A block of mass m = 2.0 kg slides head on into a spring of spring constant k = 260 N/m. When the block stops, it has compressed

the spring by 14 cm. The coefficient of kinetic friction between block and floor is 0.42. While the block is in contact with the spring and being brought to rest, what are
(a) the work done by the spring force and
(b) the increase in thermal energy of the block-floor system?
(c) What is the block's speed just as the block reaches the spring? Please show all steps.
Physics
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Explanation:

a) The work done by the spring force  

Ws =  ½ * k * x²

Ws =  ½ * 260 N/m *0.14² m  

Ws =  2.548J

b) The increase in thermal energy can by find using  

Et = Wf

Wf = µ * m *g * x  

Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m

Wf = 1.153 J

c) The speed just as the block reaches can by fin using

EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s

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Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

\Delta t = t_{A}-t_{C} (1)

Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}}  \right)

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  } (2)

Where:

v_{C} - Speed of the sound in concrete, in meters per second.

v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

If we know that \Delta t = 6.4\,s, v_{C} = 3000\,\frac{m}{s} and v_{A} = 343\,\frac{m}{s}, then the distance travelled by the sound is:

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

The impact occured at a distance of 2478.585 meters from the person.

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When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f
Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (r), measured in kilometers, can be represented by a right triangle:

r = \sqrt{x^{2}+y^{2}} (1)

Where:

x - Horizontal distance between the rocket and the observer, measured in kilometers.

y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

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Where:

\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

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If we know that y = 4\,km and \dot y = 400\,\frac{km}{h}, then the rate of change of the angle of elevation is:

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Explanation:

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5.0 kg × Velocity of the dog = 90 kg·m/s

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