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ra1l [238]
3 years ago
6

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi

ble friction about a stationary horizontal axis. We tie the free end of the cable to a 13.0 kg block and release the block from rest. As the block falls, the cable unwinds without stretching or slipping. How far will the mass have to descend to give the cylinder 510 J of kinetic energy?
Physics
1 answer:
antoniya [11.8K]3 years ago
3 0

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

Kc=510=1/2*Ic*\omega c^2

\omega c=\sqrt{510*2/Ic}

Where the inertia is given by:

Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2

Replacing this value:

\omega c=106.46rad/s

Speed of the block will therefore be:

V_b=\omega_c*R_c=106.46*0.15=15.969m/s

By conservation of energy:

Eo = Ef

Eo = 0

Ef = 510+1/2*m_b*V_b^2-m_b*g*h

So,

0 = 510+1/2*m_b*V_b^2-m_b*g*h

Solving for h we get:

h=16.67m

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Answer:

  • a) See explanation below

  • b) At X.

Explanation:

Please, see the picture attached with the image of the plastic bottle for this question.

<u>(a) Explain why the water could not flow out of the bottle.</u>

What makes the water flow out of the botlle is the force of gravity, whic attracts the water towards the Earth.

When Johnson made a small hole at the bottom of the plastic bottle containing water, the air outside the bottle, which surrounds it and exerts a pressure all over the outer walls of the bottle, exerted a force against the small area of water "over" the hole that is in contact with the air.

Thus, this force of the air pushing upward through the wall opposed the force of gravity pulling downward making the net force zero and the water cannot fall.

<u>(b) To make the water flow out more easily, his teacher suggested making another hole. At which position - X, Y or Z, should he make the 2nd hole in order for the water to flow out the fastest?</u>

You must open the hole at a place where there is not water but air, such that the outer air can enter in the bottle.

That will make that the pressure in the space over the water inside the bottle be equal to the pressure outside.

The pressure of the air above the water will push it downward. Now, the force from the pressure of air inside the water, which is downward, opposes the upward force from the pressure of air around the first hole, and the net force will be downward, making the water flow out more easily.

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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

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