Work and Energy Question? Help

In summary: Leright is trying to help you understand the problem better. The net work of 1200J is converted into kinetic energy as the crate is pulled 40m. This means that the crate will have a certain velocity at 40m. To find this velocity, you can use the equation KE=1/2mv^2, where KE is the kinetic energy, m is the mass of the crate, and v is the velocity. So, in summary, the net work of 1200J is converted into kinetic energy, which can be used to calculate the velocity of the crate at 40m.
  • #1
chesshaha
14
0
Work and Energy Question?? PLZ Help!



Hi, I am having a stupid question with a classic work and engergy problem.

Suppose you have A 50-kg crate is pulled 40m along a horizontal floor by a constant force exerted by a person, F done of by the pulling force is 100N, which at 37 degrees. F done by the friction is 50N.

I calculated the work done by the pulling force, which is 3200J.
I also calculated the work done by the frictional force, which is -2000J.

My question is what happened to the net work, 1200J.
Does the energy stores inside the crate, what?

PLZ HELP! :yuck: :cry: :confused: :tongue2: :wink:
 
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  • #2
Please post in the homework section.
 
  • #3
I need to see some work if I am to help you.

Do use the homework help section next time.
 
Last edited:
  • #4
The net work done on an object is the total kinetic energy of the object...this is the work-KE theorem. So, in moving your 50 kg mass 40m with a net work of 1200J, you add 1200J of KE to the object. From this you can calculate the velocity of the object at the time you reach 40m.
 
  • #5
I don't have a homework problem, and this is not a homework question, read my post please.

Im asking where did the net work, 1200J went? Did it store in the object?
 
  • #6
leright said:
The net work done on an object is the total kinetic energy of the object...this is the work-KE theorem. So, in moving your 50 kg mass 40m with a net work of 1200J, you add 1200J of KE to the object. From this you can calculate the velocity of the object at the time you reach 40m.

You basicly just re-stated my question, thanks...
 
  • #7
I don't think that's a restatement of your problem at all. You don't mention velocity for one thing.

What exactly is the problem you are tring to solve? What do you have to calculate?

Read again and try what leright suggests.
 
  • #8
chesshaha said:
You basicly just re-stated my question, thanks...

Lose the attitude.
 

1. What is the definition of work in physics?

In physics, work is defined as the product of force and displacement. It is a measure of the energy transferred to or from an object by applying a force over a certain distance.

2. How is work related to energy?

Work and energy are closely related concepts. Work is the transfer of energy from one system to another, or the conversion of energy from one form to another. In other words, work is a way of measuring the changes in energy of a system.

3. What are the different types of energy?

In physics, there are several types of energy, including kinetic energy, potential energy, thermal energy, electrical energy, chemical energy, and nuclear energy. Each type of energy has its own unique characteristics and can be converted into other forms of energy.

4. How is energy conserved in a system?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that the total energy in a closed system remains constant over time.

5. How can I calculate the work done on an object?

The work done on an object can be calculated by multiplying the force applied to the object by the distance the object moves in the direction of the force. The unit for work is joules (J), which is equivalent to kg*m^2/s^2.

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