How much work is done by the chain lifting the Magnum XL-200 lifting a 10,000 kg roller coaster train up 60 m? What are you looking for?

Question

Responses

Work
Work

Kinetic Energy
Kinetic Energy

Gravity Potential Energy
Gravity Potential Energy

Spring Potential Energy

Answer 1

To calculate the work done by the chain in lifting the Magnum XL-200 roller coaster train (10,000 kg) up 60 m, we can determine the gravitational potential energy gained by the roller coaster train, which is equal to the work done against gravity.

The formula for gravitational potential energy (PE) is given by:

\[ \text{PE} = mgh \]

where:

• \( m \) is the mass (10,000 kg),
• \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and
• \( h \) is the height (60 m).

Using these values:

\[ \text{PE} = 10,000 , \text{kg} \times 9.81 , \text{m/s}^2 \times 60 , \text{m} \]

\[ \text{PE} = 10,000 \times 9.81 \times 60 \]

\[ \text{PE} = 5886000 , \text{Joules} \]

Therefore, the work done by the chain in lifting the roller coaster train is 5,886,000 Joules (or 5.886 MJ).

In this context, the correct response is Gravity Potential Energy.

Answer 2

How much work is done by the chain lifting the Magnum XL-200 lifting a 10,000 kg roller coaster train up 60 m? Which of these things are ignored for this problem?
Responses

Work
Work

Kinetic Energy
Kinetic Energy

Gravity Potential Energy
Gravity Potential Energy

Spring Potential Energy

Answer 3

In the context of the problem, we are specifically calculating the work done by the chain in lifting the roller coaster train, which relates to gravitational potential energy. Here’s a breakdown:

1. Work: This is what we are calculating.
2. Kinetic Energy: This is not relevant because the problem involves lifting the train, not its motion. We are not considering the kinetic energy of the moving train while being lifted.
3. Gravity Potential Energy: This represents the change in potential energy as the train is lifted and is integral to the calculation of work done against gravity.
4. Spring Potential Energy: This is not relevant in this scenario since we are not dealing with springs or any spring-like systems.

Given these points, the correct response to which of these things are ignored for this problem is:

Kinetic Energy and Spring Potential Energy.

Both are not significant to the work done by the chain in lifting the train in this specific context.

Answer 4

How much work is done by the chain lifting the Magnum XL-200 lifting a 10,000 kg roller coaster train up 60 m? The work done is _______ J.

Answer 5

To calculate the work done by the chain in lifting the Magnum XL-200 roller coaster train, we apply the formula for gravitational potential energy (which is equal to the work done against gravity):

\[ \text{Work} = mgh \]

Where:

• \( m = 10,000 , \text{kg} \) (mass of the roller coaster train),
• \( g \approx 9.81 , \text{m/s}^2 \) (acceleration due to gravity),
• \( h = 60 , \text{m} \) (height).

Now substituting the values into the formula:

\[ \text{Work} = 10,000 , \text{kg} \times 9.81 , \text{m/s}^2 \times 60 , \text{m} \]

Calculating this:

\[ \text{Work} = 10,000 \times 9.81 \times 60 \]

\[ \text{Work} = 5886000 , \text{J} \]

Thus, the work done by the chain lifting the roller coaster train is 5,886,000 Joules (J).

So, the complete sentence is:

The work done is 5886000 J.

Answer 6

On the Ford class of aircraft carrier has a 92 m long catapult. At launch an F/A – 18 super hornet has a launch weight of about 22,000 kg. It must go 76 m/s to get airborne at the bow of the carrier where the catapult ends. How much net force is applied to the super hornet on launch? What are you looking for in the problem?

Responses

Distance / Displacement
Distance / Displacement

Force
Force

Height
Height

Mass
Mass

Velocity
Velocity

Spring Constant
Spring Constant

Answer 7

To find the net force applied to the F/A-18 Super Hornet during launch, we can use Newton's second law of motion, which states:

\[ F = ma \]

Where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.

First, we need to determine the acceleration \( a \) required for the F/A-18 to reach its launch speed of \( 76 , \text{m/s} \) while being launched over a distance of \( 92 , \text{m} \) (the length of the catapult).

Using the kinematic equation:

\[ v^2 = u^2 + 2a s \]

Where:

• \( v = 76 , \text{m/s} \) (final velocity),
• \( u = 0 , \text{m/s} \) (initial velocity, since it starts from rest),
• \( s = 92 , \text{m} \) (distance).

Substituting the values:

\[ (76 , \text{m/s})^2 = (0 , \text{m/s})^2 + 2 a (92 , \text{m}) \]

This simplifies to:

\[ 5776 = 2 a (92) \]

\[ 5776 = 184a \]

Now, solving for \( a \):

\[ a = \frac{5776}{184} \approx 31.39 , \text{m/s}^2 \]

Now that we have the acceleration, we can calculate the net force \( F \):

\[ F = ma = 22000 , \text{kg} \times 31.39 , \text{m/s}^2 \]

Calculating the force:

\[ F \approx 691580 , \text{N} \]

Thus, the net force applied to the Super Hornet on launch is approximately 691,580 N.

In terms of what we are looking for in the problem, the correct response would be:

Force.

Answer 8

On the Ford class of aircraft carrier has a 92 m long catapult. At launch an F/A – 18 super hornet has a launch weight of about 22,000 kg. It must go 76 m/s to get airborne at the bow of the carrier where the catapult ends. How much net force is applied to the super hornet on launch? Which part of the conservation of energy has the force in it? What piece are you looking for?
Responses

Work
Work

Kinetic Energy
Kinetic Energy

Gravity Potential Energy
Gravity Potential Energy

Spring Potential Energy

Answer 9

In the context of this problem, the net force applied to the F/A-18 Super Hornet during launch is directly related to the concept of work, as work is done by a force over a distance.

The relevant relationship is that the work done by the net force is equal to the change in kinetic energy of the object. The work-energy principle states that:

\[ \text{Work} = \Delta KE \]

Where:

• Work is the energy transferred to or from an object via the application of force along a displacement.
• \( \Delta KE \) is the change in kinetic energy of the object.

Given that work involves the application of force, the correct response to your question is:

Work.

This is the part of the conservation of energy that relates to the force applied during the launch, as it corresponds to the energy transferred to the F/A-18 Super Hornet to achieve the necessary kinetic energy for takeoff.

Answer 10

On the Ford class of aircraft carrier has a 92 m long catapult. At launch an F/A – 18 super hornet has a launch weight of about 22,000 kg. It must go 76 m/s to get airborne at the bow of the carrier where the catapult ends. How much net force is applied to the super hornet on launch? The force required to launch the super hornet is _______ N. Round to the whole number.

Answer 11

To calculate the net force applied to the F/A-18 Super Hornet during launch, we first need to find the acceleration required for the aircraft to reach its launch speed of \(76 , \text{m/s}\) over the \(92 , \text{m}\) length of the catapult.

Using the kinematic equation:

\[ v^2 = u^2 + 2as \]

Where:

• \(v = 76 , \text{m/s}\) (final velocity),
• \(u = 0 , \text{m/s}\) (initial velocity),
• \(s = 92 , \text{m}\) (distance).

Substituting the values:

\[ (76 , \text{m/s})^2 = (0 , \text{m/s})^2 + 2a(92 , \text{m}) \]

This simplifies to:

\[ 5776 = 184a \]

Now solving for \(a\):

\[ a = \frac{5776}{184} \approx 31.39 , \text{m/s}^2 \]

Next, we can find the net force \(F\) using Newton's second law:

\[ F = ma \]

Where:

• \(m = 22000 , \text{kg}\) (mass of the Super Hornet),
• \(a \approx 31.39 , \text{m/s}^2\) (acceleration).

Now substituting the values into the force equation:

\[ F = 22000 , \text{kg} \times 31.39 , \text{m/s}^2 \]

Calculating the force:

\[ F \approx 691580 , \text{N} \]

Rounded to the whole number, the force required to launch the Super Hornet is:

691580 N.