Work energy principle and power

In summary, a child of mass 40kg slides down a playground slide from rest at the top, 2m above the ground. At the bottom, the slope levels off and there is a constant resistance of 112N. The child's loss of gravitational potential energy is 800J and using the work energy principle, the distance traveled when the child comes to rest is 7.14m. On the level part of the slide, the increase in kinetic energy is 0J, the increase in gravitational potential energy is -800J, and the work done by gravity is 200J. Therefore, the distance traveled on the level part is 3.14m.
  • #1
Shah 72
MHB
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0
A child of mass 40kg slides down a playground slide. The child starts from rest at the top of the slide, 2m above the ground. At the bottom of the slide it's slope levels off.
a) Find the child's loss of GPE
I got the ans 800J
there is a constant resistance of 112N throughout
b) find the distance the child has traveled when she comes to rest.
Using work energy principle
Increase in KE =0J
Increase in GPE= -800J
Work done against resistance = -112SJ
So I get S= 7.14m
The slide is inclined at an angle of 30 degree to the horizontal.
C) Find the distance the child travels on the level part of the slide.
I don't understand this part.
Increase in KE= 0J
Increase in GPE= 0- 40×10sin30×2= -400J
Work done by gravity= F×S
I don't understand how to solve further. Pls help
 
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  • #2
Shah 72 said:
A child of mass 40kg slides down a playground slide. The child starts from rest at the top of the slide, 2m above the ground. At the bottom of the slide it's slope levels off.
a) Find the child's loss of GPE
I got the ans 800J
there is a constant resistance of 112N throughout
b) find the distance the child has traveled when she comes to rest.
Using work energy principle
Increase in KE =0J
Increase in GPE= -800J
Work done against resistance = -112SJ
So I get S= 7.14m
The slide is inclined at an angle of 30 degree to the horizontal.
C) Find the distance the child travels on the level part of the slide.
I don't understand this part.
Increase in KE= 0J
Increase in GPE= 0- 40×10sin30×2= -400J
Work done by gravity= F×S
I don't understand how to solve further. Pls help
I think I did mistake while calculating Increase in GPE, it will be -800J
Work done by gravity = 200 SJ
So I get S= 4m
I subtract from full length 7.14-4= 3.14m
 

What is the work energy principle?

The work energy principle states that the work done on an object is equal to the change in its kinetic energy. In other words, the net work done on an object will result in a change in its speed or direction.

How is the work energy principle related to power?

Power is the rate at which work is done. The work energy principle is closely related to power because it shows that the amount of work done on an object is directly proportional to the power used to do that work.

What are some real-life examples of the work energy principle?

Some examples of the work energy principle in action include a person pushing a shopping cart, a car accelerating on a highway, and a roller coaster moving up and down a track. In each of these scenarios, work is being done on the object, resulting in a change in its kinetic energy.

How is the work energy principle used in engineering and physics?

The work energy principle is a fundamental concept in engineering and physics. It is used to analyze and design systems involving motion and energy, such as machines, vehicles, and structures. It allows engineers and physicists to calculate the amount of work and power needed to achieve a desired outcome and to optimize the efficiency of these systems.

What are some common misconceptions about the work energy principle?

One common misconception is that the work energy principle only applies to objects in motion. In reality, it applies to any type of work being done, including work done on stationary objects. Another misconception is that the work energy principle only applies to mechanical systems. However, it can also be applied to other forms of energy, such as thermal and electrical energy.

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