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

In summary, after pushing the object off the edge of the table, it slid across the table and then fell to the ground. The object had a speed of .935 m/s when it left the table.
  • #1
notagenius08
11
0

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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  • #2
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.
 
  • #3
Finally solved it! Thanks for your help. Still don't understand why the book listed it under nonuniform circular motion...


Thanks again.
 

1. What is nonuniform circular motion?

Nonuniform circular motion is the movement of an object along a circular path at varying speeds. This means that the object's velocity, or speed and direction, is constantly changing as it moves along the circular path.

2. What is the difference between uniform and nonuniform circular motion?

Uniform circular motion is when an object moves along a circular path at a constant speed, meaning its velocity remains the same throughout. Nonuniform circular motion, on the other hand, involves changing velocity, making the speed and direction of the object's movement constantly changing.

3. What causes nonuniform circular motion?

Nonuniform circular motion is caused by a force acting on an object that is moving along a circular path. This force, called centripetal force, pulls the object towards the center of the circle, causing it to change direction and velocity.

4. How is nonuniform circular motion measured?

Nonuniform circular motion can be measured using various mathematical formulas, such as centripetal acceleration, tangential acceleration, and angular velocity. These measurements can help determine the object's speed, direction, and acceleration as it moves along the circular path.

5. What are some real-life examples of nonuniform circular motion?

Some common examples of nonuniform circular motion include a car driving around a curved road, a satellite orbiting the Earth, and a rollercoaster moving along its track. Essentially, any time an object moves along a circular path at varying speeds, it is experiencing nonuniform circular motion.

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