Energy Calculations please check work (thanks)

In summary, the work done by the woman in lifting the barbell will be the same whether she does it in 5.0s or 10s. In the second problem, the height above the Earth's surface at which the gravitational potential energy of the earth-block system is increased by 500J is 45m, taking into account the starting height of 20m.
  • #1
blackout85
28
1
Please someone check to see if on the right track:

A woman lifts a barbell 2.0m in 5.0s, beginning and ending at rest. If she lifts the same distance in 10s, the work done by her is:

I said it would be four times as much because I used 1/2mv^2. I placed the v I calculated from v=d/t into the kinetic equation. (.4)^2 (.2)^2
.16 is four times as much as .04. Is this right.

A 2kg block is thrown upward from a point 20m above the Earth's surface. At what height above the Earth's surface will the gravitational potential energy of the earth-block systme have increased by 500J.

work:

mgh= 2(9.81)(20m)= 392.4 --> potential
500J= mgh
500J = (2kg)(9.81) (h)
h= 25m

If someone could point me in the right direction of how to do this problem I would appreciate it. Thank you.
 
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  • #2
blackout85 said:
Please someone check to see if on the right track:

A woman lifts a barbell 2.0m in 5.0s, beginning and ending at rest. If she lifts the same distance in 10s, the work done by her is:

I said it would be four times as much because I used 1/2mv^2. I placed the v I calculated from v=d/t into the kinetic equation. (.4)^2 (.2)^2
.16 is four times as much as .04. Is this right.

I don't think so - I think the work will be the same - Work Done = Force x distance - taking longer means less power but the same work.

A 2kg block is thrown upward from a point 20m above the Earth's surface. At what height above the Earth's surface will the gravitational potential energy of the earth-block systme have increased by 500J.

work:

mgh= 2(9.81)(20m)= 392.4 --> potential
500J= mgh
500J = (2kg)(9.81) (h)
h= 25m

If someone could point me in the right direction of how to do this problem I would appreciate it. Thank you.

This one I think is correct - except the actual height is 45, you need to add the 20m you started with.. When you're as close to the Earth as this, the mgh formula is fine.
 
  • #3


To check if you are on the right track, let's break down the problem into smaller steps and use the correct equations for each step.

First, let's calculate the work done by the woman in the first scenario. We can use the equation W = Fd, where W is work, F is force, and d is distance. We know that the woman lifted the barbell 2.0m, so we need to find the force she used. We can use the equation F = ma, where m is the mass of the barbell and a is the acceleration. Since the barbell was initially at rest and ended at rest, the acceleration is equal to 0. Therefore, the force used is also equal to 0. This means that the work done in the first scenario is also equal to 0.

Next, let's calculate the work done in the second scenario. We already know that the distance is 2.0m, so we just need to find the force used. Using the same equation as before, F = ma, we can find that the force used is equal to 1/2mv^2. Plugging in the values of 2kg for mass and 0.4m/s for velocity (calculated by dividing 2m by 5s), we get a force of 0.4N. Now, we can calculate the work done using the equation W = Fd. Plugging in the values of 0.4N for force and 2.0m for distance, we get a work of 0.8J.

To check if this answer is four times as much as the work done in the first scenario, we can compare it to the work done in the first scenario (which we found to be 0J). Since 0.8J is four times greater than 0J, this means that your answer is correct!

For the second problem, we are given the gravitational potential energy of the earth-block system (500J) and asked to find the height at which this energy is reached. To do this, we can use the equation for gravitational potential energy, U = mgh, where U is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.

We already know the values for m (2kg) and g (9.81m/s^2), so we can rearrange the equation to solve
 

What are energy calculations?

Energy calculations are mathematical calculations used to determine the amount of energy needed or produced in a system. This can include calculating the potential energy, kinetic energy, or total energy of an object or system.

What units are used in energy calculations?

The most common units used in energy calculations are joules (J) and kilojoules (kJ). However, other units such as calories, electron volts, and British thermal units (BTUs) may also be used depending on the specific application.

How do you calculate potential energy?

Potential energy can be calculated by multiplying the mass of an object by the acceleration due to gravity (9.8 m/s^2) and the height at which the object is located (PE = mgh). This equation assumes that the object is near the surface of the Earth and is not moving.

What is the difference between potential energy and kinetic energy?

Potential energy is the energy an object possesses due to its position or state, while kinetic energy is the energy an object possesses due to its motion. Potential energy can be converted into kinetic energy and vice versa.

How can energy calculations be applied in real-life situations?

Energy calculations are used in a variety of real-life situations, such as determining the amount of energy needed to power a home or calculating the potential energy of a roller coaster. They are also essential in fields such as engineering, physics, and chemistry for designing and analyzing systems and processes.

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