What was the object's speed as it fell off the edge of the table?

AI Thread Summary
The discussion revolves around calculating the speed of a 100 g object as it falls off a table after being pushed. The object slides 1.60 m before falling 0.900 m and landing 40.0 cm away from the table edge, with a coefficient of kinetic friction of 0.300. The initial velocity calculated was 0.935 m/s, but there was an error in calculating the friction force due to incorrect unit conversion. The solution involves using the work-energy theorem or Newton's second law along with kinematic equations. The participant ultimately solved the problem but questioned its classification under nonuniform circular motion.
notagenius08
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Homework Statement



This homework problem is under the Nonuniform circular motion problem, but seems more like a elementary kinematics.

You push the 100 g object and release it 1.60 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 0.900 m to the floor, and lands 40.0 cm from the edge of the table.

If the coefficient of kinetic friction is 0.300, what was the object's speed as you released it?

Homework Equations


t=sqrt of 2*height/9
Fk=coeff of friction*m*a =n*coeff of fric.

The Attempt at a Solution



Figured velocity as object leaves table to be .935 m/s
and Fk=295N

I am stuck on what to do next, can't find v,a, or t.
 
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notagenius08 said:

Homework Statement



This homework problem is under the Nonuniform circular motion problem, but seems more like a elementary kinematics.

You push the 100 g object and release it 1.60 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 0.900 m to the floor, and lands 40.0 cm from the edge of the table.

If the coefficient of kinetic friction is 0.300, what was the object's speed as you released it?


Homework Equations


t=sqrt of 2*height/9
Fk=coeff of friction*m*a =n*coeff of fric.


The Attempt at a Solution



Figured velocity as object leaves table to be .935 m/s
and Fk=295N

I am stuck on what to do next, can't find v,a, or t.
Your velocity as the object leaves the table looks good, but you made an error in determining Fk, you forgot to convert grams to kg (100grams = 0.1kg). Now once you correct the value for the friction force, ther are a number of ways to determine the initial speed of the object at its release, by looking at the forces acting on the block after its release, while it is on the table. The easiest way is the use of the work-energy theorem, if you are familiar with that method; or else, Newton 2 and the kinematic equations will do it. Give it a try.
 
Finally solved it! Thanks for your help. Still don't understand why the book listed it under nonuniform circular motion...


Thanks again.
 
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