Why is the acceleration not equal to 9.81 m/s^2

In summary: You misunderstood me. I meant that the weight of the 50 g mass divided by the acceleration will not result in the mass of the puck.
  • #1
rexorsist
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Homework Statement



I'm doing a lab where there is a pulley system. A mass of 50g is attached to a string that goes through the pulley, which is then attached to a puck, on a frictionless air table. So when the mass is let go, the 50g pulls down the puck to the edge of the table.

So why isn't the acceleration of the puck = 9.81 m/s^2, knowing that the 50g mass is in free fall, pulling the puck?

I got an average acceleration of 33.69 cm/s^2.

Would I be correct to say that it's because the force of gravity would equal the acceleration of the puck? So I would multiply 0.05 kg by 981 cm/s^2 to get an answer of 49.05 cm/s (which would be the acceleration I should've gotten, instead of 33.69 cm/s^2). Is this at all correct? ...or am I completely off?

Thanks in advance!
 
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  • #2
You forgot to state the mass of the puck and the pulley.


The reason you will get a lower acceleration than g is that gravity only acts on the 50g mass to accelerate the system. But the total inertia of the system is the 50g mass plus the mass of the puck plus the mass of the pulley.
 
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  • #3
Thanks a lot!

But would I be correct to say that in order to find the mass of the puck, I could simply multiply 0.05 kg by 9.8 m/s^2 to get 0.49 N as a force, then divide it by my acceleration.

Or would that be the incorrect force?

My teacher is forcing us to find the mass by creating our own equation based on the pulley system.

Would my method be valid as well?
 
  • #4
There is no difference between your method and what the teacher is asking. The teacher wants you to use your method to find an equation. Pretend you don't know the hanging mass or what the local gravity is. Instead of using 50 g you use m and instead of using 9.81 you use g. Than you use algebra (which by the way is much easier than plugging in the numbers) to solve the problem.
 
  • #5
No, your method will not result in the mass of the puck.
Do as you teacher said. Write the equation (Newton's second law) for the system and you will see why. Guessing blindly is not the right way to learn physics.
 
  • #6
Your method is not valid.

Start by doing a free body diagram for each mass and set up two equations of motion.
 
  • #7
nasu said:
No, your method will not result in the mass of the puck.
Do as you teacher said. Write the equation (Newton's second law) for the system and you will see why. Guessing blindly is not the right way to learn physics.

How am I guessing blindly? I'm asking a question in order to correct a possible mistake, which is pretty natural when learning almost anything. Rather than telling me to see for myself, you can explain a concept I don't understand, because I've made the equation and still don't know why my method doesn't work (which I know is incorrect, but I don't understand why the force of gravity can't be applied as the net force of the puck).
 
  • #8
Well, you did not show the free body diagram or any equation.
You just asked if your series of arithmetical operations will provide the mass of the puck.
I answered that question (no) and I suggested a way to obtain the right answer.

If you did the diagram you should observe that the acceleration of the puck is due to the tension in the string and not to the weight of the 50g mass.
So the weight of the 50 g mass divided by the acceleration will not result in the mass of the puck.
But maybe I misunderstood you and you meant something else.

It is a good idea to show equations rather than describing operations in words. It will make what you mean much clearer.
 

1. Why is the acceleration not exactly 9.81 m/s^2?

The acceleration due to gravity, denoted by g, is a constant value of 9.81 m/s^2 on Earth's surface. However, this value can vary slightly based on location, altitude, and other factors. Additionally, the acceleration of an object depends on the net force acting on it, which can be affected by other forces such as air resistance. Therefore, the acceleration may not always be exactly 9.81 m/s^2.

2. Why is the acceleration sometimes greater than 9.81 m/s^2?

Acceleration can be greater than 9.81 m/s^2 in situations where the net force acting on an object is greater than its weight. This can occur in scenarios such as freefall or when an object is being pushed or pulled with a greater force than the force of gravity. In these cases, the acceleration will be greater than 9.81 m/s^2.

3. Why is the acceleration sometimes less than 9.81 m/s^2?

Acceleration can be less than 9.81 m/s^2 when the net force acting on an object is less than its weight. This can happen when there is a force acting in the opposite direction of gravity, such as when an object is being slowed down or pushed upwards. In these cases, the acceleration will be less than 9.81 m/s^2.

4. Why is the acceleration different for different objects?

The acceleration of an object depends on its mass and the net force acting on it. Objects with different masses will have different accelerations even if they are subjected to the same force. For example, a lighter object will experience a greater acceleration than a heavier object with the same force applied to it. Therefore, the acceleration can vary for different objects.

5. Why is the acceleration not always constant for an object?

Acceleration is the rate of change of an object's velocity, and it can change if the net force acting on the object changes. If the force acting on an object increases or decreases, the acceleration will also change accordingly. Additionally, external factors such as air resistance or friction can also affect the acceleration of an object, causing it to vary over time. Therefore, the acceleration may not always be constant for an object.

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