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worty [1.4K]
2 years ago
15

A new ride being built at an amusement park includes a vertical drop of 70.0 meters. Starting from rest, the ride vertically dro

ps that distance before the track curves forward. The velocity at the bottom of the drop is 12.0 m/s and the mass of the cart and passengers is 3.5 x 104 kg.
a. Calculate the potential, kinetic and total energy at the top.



b. Calculate the potential, kinetic and total energy at the bottom.

c.Is the total energy at the top equal to the total energy at the bottom? Justify your observation.
Physics
1 answer:
Alexeev081 [22]2 years ago
7 0

The potential, kinetic and total energy at the top are 2.38 x 10⁷ Joule,  0 Joule, 23.8 x 10⁶ Joule respectively and the potential, kinetic and total energy at the bottom are 0 Joule, 2.448 x 10⁶ Joule, 2.448 x 10⁶ Joule  respectively. moreover the total energies are not same as not all potential energy converting into Kinetic energy of vertically drop of ride as the ride moves straight afterwards so that also need some energy which is again coming from potential energy.

Energies at the top:-

Potential energy = mgh

Potential energy  = 3.4 x 10⁴  × 10 × 70 Joule

Potential energy  = 2.38 x 10⁷ Joule

Kinetic Energy = (1/2) mv²

Kinetic Energy = (1/2) × 3.4 x 10⁴ × 0²

Kinetic Energy = 0 Joule

Total Energy = Potential Energy + Kinetic Energy

Total Energy = ( 23.8 x 10⁶ Joule ) + ( 0 Joule )

Total Energy = 23.8 x 10⁶ Joule

Energies at the bottom:-

Potential energy = mgh

Potential energy  = 3.4 x 10⁴  × 10 × 0 Joule

Potential energy  = 0 Joule

Kinetic Energy = (1/2) mv²

Kinetic Energy = (1/2) × 3.4 x 10⁴ × 12²

Kinetic Energy = 2.448 x 10⁶ Joule

Total Energy = Potential Energy + Kinetic Energy

Total Energy = ( 0 Joule ) + ( 2.448 x 10⁶ Joule )

Total Energy = 2.448 x 10⁶ Joule

The total energies are not the same since not all potential energy is converted into Kinetic energy of the vertical drop of the ride as the ride travels straight thereafter thus that also requires some energy which is again derived from potential energy.

So we have calculated the potential, kinetic and total energy at top and bottom both, and also stated the reason of why total energy at the top is not equal to the total energy at the bottom.

Learn more about Energy here:

brainly.com/question/18963960

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1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

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