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tekilochka [14]
3 years ago
9

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

ant value of 316N/m, until the spring is compressed by 35cm. The block is then released. What is the block's final speed, in m/s?
Physics
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

U=\frac{1}{2}kx^2

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

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

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s

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Basile [38]

1) Mechanical energy is conserved in all the 4 situations

2) True

3) The potential energy of the block is 490 J

Explanation:

1)

Mechanical energy is the sum of potential energy (PE) and kinetic energy (KE) of a body:

E=PE+KE

According to the law of conservation of energy, in absence of non-conservative forces (such as air resistance, friction...), the mechanical energy of a body is always conserved.

This means that the mechanical energy is conserved in all the situations described. More specifically:

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Roller Coaster: there is a continuous conversion between kinetic energy and gravitational potential energy

2)

As we  said, the mechanical energy of the object falling down at any point of the fall is

E=KE+PE

where KE is the kinetic energy and PE is the potential energy. The value of E is constant, since the mechanical energy is conserved.

The potential energy is given by

PE=mgh

where m is the mass of the object, g is the acceleration of gravity, h is the height of the object above the ground.

As the object falls, its height h decreases, and therefore, the potential energy PE also decreases. But we said that E must remain constant: therefore, if PE decreases, this means that KE increases, therefore as the object falls, the potential energy is converted into kinetic energy (in fact, the  speed of the object increases). So the statement is true.

3)

The potential energy of an object is given by the equation

PE=mgh

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object relative to the ground

For the block in this problem, we have:

m = 10 kg

h = 5 m

Substituting, we find its potential energy:

PE=(10)(9.8)(5)=490 J

Learn more about potential energy and kinetic energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

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