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3 years ago

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A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left and two weights of the same mass, on the right.

The width of each weight is 4 cm and each group of weights is placed 4 cm from the ends. Where is the center of mass of the barbell as measured from the mid-point, M, of the bar? The bar is of uniform mass and has mass 5 kg, and the retaining collars are of negligible mass. Take to the right as positivea. -5.90cmb. -11.6cmc. +13.7cmd. +5.90cme. none of the above

1 answer:

VMariaS [17]3 years ago

4 0

Answer:

b. -11.6 cm

Explanation:

We have given parameters:

Length, l = 1.5 m = 150 cm

Mass of weight, $m_1$ = 20 kg

Width, x = 4 cm

Distance d = 4 cm

Mass of bar, $m_{bar}$ = 5 kg

We are asked to find the center of mass from the mid-point, $X_{CM} = ?$

Since 3 weights are on the left and 2 weights are on the right, we know:

$m_{left}$ = 3 * 20 = 60 kg

$m_{right}$ = 2 * 20 = 40 kg

And also we know that, $M = \frac{l}{2}$ = 150/2 = 75 cm

For the left side, center of mass is:

$x_{left} = \frac{3 * 4}{2} = 6$ cm

From the midpoint, the distance to the left is:

$X_{left} = -(M - 4 - x_{left}) = -(75 - 4 -6) = -65$ cm

For the right side, center of mass is:

$x_{right} = \frac{2 * 4}{2} = 4$ cm

From the midpoint, the distance to the right will be:

$X_{right} = (M - 4 - x_{right}) = (75 - 4 - 4) = 67$ cm

Hence,

$X_{CM} = \frac{m_{right}*x_{right} + m_{left}*x_{left} }{m_{right} + m_{left} + m_{bar}} = \frac{40 * 67 - 60 * 65}{40 + 60 + 5} = -11.62$ cm

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50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

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1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

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5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

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where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

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

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

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

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

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The kinetic energy of the ball halfway through the fall is given by

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$KE=\frac{1}{2}(2)(19.8)^2=392 J$

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The kinetic energy of the ball just before hitting the ground is given by

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Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

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