Sliding Block - True or False?

In summary: Eagle's reasoning was incorrect because they assumed that work done by gravity and potential energy were the same, but in reality, potential energy is equal to the negative of work done by gravity. This means that the change in potential energy is equal to the negative of the work done by gravity, which leads to answer 2 being false. In summary, the conversation discusses the difference between work done by gravity and gravitational potential energy, with the conclusion that they are not the same and that potential energy is equal to the negative of work done by gravity. This leads to one of the options in a question being false and the correct formula for work being W=Fdcos θ.
  • #1
hb20007
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Homework Statement



It's in the picture attached...

Homework Equations



None...

The Attempt at a Solution



I picked options 2, 5 and 7 to be true but 2 turned out to be false. I can't figure out why it's wrong and an explanation would be appreciated. I mean it makes sense to me because work done by gravity is F*s which is (m*g)*h which is the same as the formula for gravitational potential energy, U = mgh
 

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  • #2
i assume that you were using W=mgh and U=mgh? this is only correct when it is vertical, but since there is an angle θ, you need to use that other Work equation (W=Fdcos θ) which means that potential energy is not the same as Work. again, that would only be true if it were vertical such as a raindrop.
 
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  • #3
Got it. Thanks
 
  • #4
Another Question !

Eagle's Wings said:
i assume that you were using W=mgh and U=mgh? this is only correct when it is vertical, but since there is an angle θ, you need to use that other Work equation (W=Fdcos θ) which means that potential energy is not the same as Work. again, that would only be true if it were vertical such as a raindrop.

I was thinking of applying this logic to a pendulum situation.

Let's say the pendulum is in motion and is at a certain moment in time at an angle of 45° to the vertical. I know that the work done by the tension is 0 as it's tangential to the direction of motion. What about the work done by gravity until the point where the pendulum is vertical?

Is it mgh or mgh*sin θ? I think it's just mgh because the equation for work done by gravity was given to me as mg(L-L*cos θ) where L = Length of string at that moment in time and it follows that (L-L*cos θ) is = h.

If it indeed is mgh, then it looks like we have assumed that the force of gravity only does work in the downwards direction. As tension does no work, what force does work in the direction of motion??
 
  • #5
hb20007 said:
I picked options 2, 5 and 7 to be true but 2 turned out to be false. I can't figure out why it's wrong and an explanation would be appreciated. I mean it makes sense to me because work done by gravity is F*s which is (m*g)*h which is the same as the formula for gravitational potential energy, U = mgh
Pay careful attention to the wording of option 2. When the block slides down, is the work done by gravity positive or negative? Is the change in potential energy positive or negative?
 
  • #6
Eagle's Wings said:
i assume that you were using W=mgh and U=mgh? this is only correct when it is vertical, but since there is an angle θ, you need to use that other Work equation (W=Fdcos θ) which means that potential energy is not the same as Work. again, that would only be true if it were vertical such as a raindrop.
Note that h = d cosθ, so the two expressions for work are equivalent.
 
  • #7
Doc Al said:
Note that h = d cosθ, so the two expressions for work are equivalent.

Oh. Does this mean that Eagle's answer was wrong and that the reason answer 2 is false is not because the formula of work is not mgh as Eagle argued but because change in potential energy is equal to - (work done by gravity) and not equal to work done by gravity itself?
 
  • #8
hb20007 said:
Oh. Does this mean that Eagle's answer was wrong and that the reason answer 2 is false is not because the formula of work is not mgh as Eagle argued but because change in potential energy is equal to - (work done by gravity) and not equal to work done by gravity itself?
Yes.
 

1. Is the "Sliding Block" experiment a true or false statement?

The "Sliding Block" experiment is a false statement. It is a classic example of a thought experiment used in physics to illustrate concepts and principles, but it does not actually represent a real physical scenario.

2. How does the "Sliding Block" experiment demonstrate Newton's Laws of Motion?

The "Sliding Block" experiment illustrates Newton's First Law of Motion, also known as the Law of Inertia. This law states that an object at rest will stay at rest and an object in motion will stay in motion unless acted upon by an external force. In the experiment, the block remains at rest until the table is pulled out, providing the external force needed to set the block in motion.

3. Can the "Sliding Block" experiment be used to explain friction?

No, the "Sliding Block" experiment does not accurately represent the concept of friction. In the experiment, the block slides without any friction, which is not how objects behave in the real world. Friction is a force that opposes motion and is essential for objects to stay in place or move in a controlled manner.

4. What is the purpose of the "Sliding Block" experiment?

The purpose of the "Sliding Block" experiment is to demonstrate the concept of inertia and how external forces can affect the motion of an object. It is also used to introduce the concept of Newton's First Law of Motion and to help students understand the difference between theoretical and real-world scenarios.

5. Can the "Sliding Block" experiment be applied to other real-world situations?

While the "Sliding Block" experiment is not an accurate representation of a real-world scenario, the concepts and principles it illustrates can be applied to various situations, such as the motion of objects on an inclined plane or the movement of a car on a level road. However, it is important to note that these situations involve additional factors such as friction, air resistance, and external forces, which are not accounted for in the experiment.

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