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3 years ago

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A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-

the loop which has a radius 5 m. The acceleration of gravity is 9.8 m/s^2.
Required:
How large is the normal force on it if its mass is 5g?

1 answer:

solong [7]3 years ago

3 0

Answer:

N₁ = 393.96 N and N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

N -W = m a

where the acceleration is ccentripeta

a = v² / r

N = W + m v² / r

N = mg + mv² / r

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

Em₀ = U = m g h

lowest point. Stop curl down

$Em_{f}$ = K = ½ m v²

Emo = Em_{f}

m g h = ½ m v²

v² = 2 gh

we substitute

N = m (g + 2gh / r)

N = mg (1 + 2h / r)

let's calculate

N₁ = 5 9.8 (1 + 2 17.6 / 5)

N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

-N -W = - m v₂² / r

N = m v₂² / r - W

N = m (v₂²/r - g)

we seek speed with the conservation of energy

Em₀ = U = m g h

final point. Top of circle with height 2r

$Em_{f}$ = K + U = ½ m v₂² + mg (2r)

Em₀ = Em_{f}

mgh = ½ m v₂² + 2mgr

v₂² = 2 g (h-2r)

we substitute

N = m (2g (h-2r) / r - g)

N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)

N = mg (2h/r - 3)

N = 5 9.8 (2 17.6 / 5 -3)

N = 197.96 N

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3 years ago

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

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Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$

The expression is now expanded:

$E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}$

Where $v_{x}$ is the horizontal speed of the arrow, measured in meters per second.

$v_{x} = v_{1}\cdot \cos \theta$

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the horizontal speed is:

$v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}$

$v_{x} \approx 74.911\,\frac{m}{s}$

If $m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{max} = 56.712\,m$ and $v_{x} \approx 74.911\,\frac{m}{s}$ , the total mechanical energy is:

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}$

$E = 201.720\,J$

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

$E = m\cdot g \cdot y_{max}$

$m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $y_{max} = 342.816\,m$

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)$

$E = 201.720\,J$

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0

3 years ago

The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10m/s^2, the range of projectile in meters is?

Morgarella [4.7K]

Answer:

18 m

Explanation:

y = 10x − 5/9 x²

When the projectile lands, y=0:

0 = 10x − 5/9 x²

0 = x (10 − 5/9 x)

x = 0, 18

The range of the projectile is 18 m.

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