Help needed on Kinetic Energy and Work Problem

In summary: The normal force is internal to the system and acts on both cab and cheese, but in opposite directions.
  • #1
ctpengage
33
0

Homework Statement



A 0.250 kg block of cheese lies on the floor of a 900 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.40 m and then through distance d2 = 10.5 m. (a) Through d1, if the normal force on the block from the floor has constant magnitude FN = 3.00 N, how much work is done on the cab by the force from the cable? (b) Through d2, if the work done on the cab by the (constant) force from the cable is 92.61 kJ, what is the magnitude of FN?

2. The attempt at a solution

Solution Part A

The net upward force is given by

FTension + FN - (m+M)g = (m+M)a
where m = 0.250 kg is the mass of the cheese, M = 900 kg is the mass of the elevator cab,
F is the force from the cable, and 3.00 N N F = is the normal force on the cheese. On the
cheese alone, we have

FN - mg = ma
Therefore a = [3.00-(0.250)(9.80)] / 0.250 = 2.20 m/s2

Thus the force from the cable is F FTension = (m+M)(a+g) - FN = 1.08 x 104 and the work done by the cable on the cab is

W = FTensiond1 = (1.08 x 104)(2.40) = 2.59 x 104[/SUP

However I am stuck on Part B. The answer for Part B is 2.45 N. Can anyone please help me on Part B? Thanks guys.
 
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  • #2
Anyone please help?
 
  • #3
ctpengage said:
The net upward force is given by

FTension + FN - (m+M)g = (m+M)a
where m = 0.250 kg is the mass of the cheese, M = 900 kg is the mass of the elevator cab,
F is the force from the cable, and 3.00 N N F = is the normal force on the cheese.
The net force on the entire system (cab + cheese) would not include the normal force. The normal force is internal to the system and acts on both cab and cheese, but in opposite directions.

On the
cheese alone, we have

FN - mg = ma
Therefore a = [3.00-(0.250)(9.80)] / 0.250 = 2.20 m/s2
Good.

Thus the force from the cable is F FTension = (m+M)(a+g) - FN = 1.08 x 104 and the work done by the cable on the cab is

W = FTensiond1 = (1.08 x 104)(2.40) = 2.59 x 104
Redo this calculation using the correct net force.

However I am stuck on Part B. The answer for Part B is 2.45 N. Can anyone please help me on Part B? Thanks guys.
Start by finding the cable tension. Then just work backwards, using the same approach as above.
 
  • #4
Doc Al said:
The net force on the entire system (cab + cheese) would not include the normal force. The normal force is internal to the system and acts on both cab and cheese, but in opposite directions.


Good.


Redo this calculation using the correct net force.


Start by finding the cable tension. Then just work backwards, using the same approach as above.


Why was my net force wrong?
 
  • #5
ctpengage said:
Why was my net force wrong?
See my last post:
Doc Al said:
The net force on the entire system (cab + cheese) would not include the normal force.
 

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is a type of energy that is dependent on an object's mass and velocity.

2. How is kinetic energy calculated?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. What is the relationship between kinetic energy and work?

Work and kinetic energy are closely related, as work done on an object causes a change in its kinetic energy. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

4. Can you provide an example of a kinetic energy and work problem?

Sure, let's say a car with a mass of 1000 kg is traveling at a speed of 20 m/s. What is its kinetic energy? To solve this problem, we use the formula KE = 1/2 * m * v^2. Plugging in the values, we get KE = 1/2 * 1000 kg * (20 m/s)^2 = 200,000 joules. Now, let's say the car accelerates to a speed of 30 m/s. How much work was done on the car? To find the work, we use the formula W = KEf - KEi, where KEf is the final kinetic energy and KEi is the initial kinetic energy. So, the work done on the car is W = (1/2 * 1000 kg * (30 m/s)^2) - (1/2 * 1000 kg * (20 m/s)^2) = 400,000 joules.

5. What are some real-world applications of kinetic energy and work?

Kinetic energy and work are important concepts in many fields, including physics, engineering, and sports. They are used in designing efficient machines and structures, calculating the energy required for different movements, and predicting the motion of objects in different scenarios. Some examples of real-world applications include calculating the energy needed to propel a rocket into space, designing roller coasters, and understanding the mechanics of athletic movements in sports such as running, jumping, and throwing.

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