Work - Energy Theorem Problem

In summary: I took cos instead of sin. Thank you for clarifying that for me!In summary, the problem involves a block of ice sliding down an inclined plane with an angle of 36.9° below the horizontal. The block starts from rest and the goal is to find its final speed. Using the equation Wtot = 1/2mv^2 (final) - 1/2mv^2 (initial), the work total was calculated to be 11.75J. Then, using the formula mgh=.5mv^2, the final speed was found to be 3.42 m/s. However, the correct solution uses the formula KE_{i}+PE_{i}=KE_{f}+PE
  • #1
rodrj183
4
0

Homework Statement



A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9° below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

This problem is found in the University Physics With Modern Physics 13th Edition Book. It is problem number 6.28 in Chapter 6.

Homework Equations



Wtot = Ʃ F times d

Wtot = 1/2mv^2 (final K2) - 1/2 mv^2 (initial K1)

The Attempt at a Solution



Hello guys! I hope everyone is well! There is something I'm not really quite understanding about this

particular problem. Heres my attempt at the solution:

First I found the Work Total done.

I went ahead and found the parallel component of the ice blocks weight to the slope and multiplied it

by the distance of 0.750m down the slope. This gave me a Work Total of 11.75J

I then used the formula Wtot = 1/2mv^2 (final) - 1/2mv^2 (initial) with K1 being 0 since the ice block starts from rest.

11.75J = 1/2(2kg)v^2 - 0

divided 11.75J by 1kg to give me 11.75 m^2/s^2

took sqrt of 11.75 m^2/s^2

V=√11.75 m^2/s^2 = 3.42 m/s

The answer in the book is 2.97m/s.

Im not sure where I went wrong exactly, and I am having a hard time trying to figure out the correct solution. I would love some help! Thanks!
 
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  • #2
work done by gravity = change in kinetic energy...!
mgh=.5mv2
h is height of the wedge...!
 
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  • #3
Just use the fact that:
[tex]KE_{i}+PE_{i}=KE_{f}+PE_{f} \implies \frac{1}{2}mv^{2}_{i}+mgh_{i}=\frac{1}{2}mv^{2}_{f}+mgh_{f}[/tex]
The initial KE and final PE are going to be what?? Then solve from there.

Edit: The trick here is the h isn't the length it slides. Draw a triangle with the hypotenuse of the length it slides and theta being the angle given, then find the height in the y direction.
 
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  • #4
I got the answer now, but how come the component of weight parallel to the slope wasn't a valid way of finding the solution? Is it because gravity was the only force doing the work, and gravity can only displace vertically up and down?
 
  • #5
iRaid said:
Just use the fact that:
[tex]KE_{i}+PE_{i}=KE_{f}+PE_{f} \implies \frac{1}{2}mv^{2}_{i}+mgh_{i}=\frac{1}{2}mv^{2}_{f}+mgh_{f}[/tex]
The initial KE and final PE are going to be what?? Then solve from there.

Edit: The trick here is the h isn't the length it slides. Draw a triangle with the hypotenuse of the length it slides and theta being the angle given, then find the height in the y direction.

I just actually realized something. The chapter that's in my book doesn't cover GPE yet so the only formulas I was introduced to was the following. The book never mentioned anything about MGH in this chapter. The next chapter mentions it.

Work = Fs cosθ

Wtot = Ʃ F s

and

Wtot = 1/2mv^2 (final) - 1/2mv^2 (initial)
 
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  • #6
rodrj183 said:
I got the answer now, but how come the component of weight parallel to the slope wasn't a valid way of finding the solution?
Your method was valid, but your numbers look wrong. Did you take cos instead of sine?
 
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  • #7
haruspex said:
Your method was valid, but your numbers look wrong. Did you take cos instead of sine?

Yep that's exactly what I did.
 

1. What is the Work-Energy Theorem?

The Work-Energy Theorem is a physics principle that states that the work done on an object is equal to the change in its kinetic energy. This means that the net work done on an object will result in a change in its speed or direction of motion.

2. How is the Work-Energy Theorem used in problem solving?

The Work-Energy Theorem is often used to solve problems involving the motion of objects. By calculating the work done on an object and its initial and final kinetic energies, the theorem can be applied to find the object's final speed or displacement.

3. Can the Work-Energy Theorem be applied to all types of motion?

Yes, the Work-Energy Theorem can be applied to all types of motion, including linear, rotational, and oscillatory motion. As long as there is a change in kinetic energy, the theorem can be used to analyze the motion of an object.

4. How is the Work-Energy Theorem related to the Law of Conservation of Energy?

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred or transformed. The Work-Energy Theorem is a specific application of this principle, as it shows how work done on an object is equal to the change in its kinetic energy, demonstrating the conservation of energy.

5. What are some real-world applications of the Work-Energy Theorem?

The Work-Energy Theorem has many practical applications, such as calculating the efficiency of machines, determining the speed of moving objects, and understanding the forces acting on a roller coaster or car during its motion. It is also used in fields such as engineering, sports, and transportation to analyze and optimize various systems and movements.

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