2D Kinematics Problem w/ Coefficient of Friction?

In summary, the conversation discusses a physics problem involving a sliding block on a frictionless lab bench. The block hits a rough patch before sliding off the bench and the goal is to find the coefficient of friction between the block and the rough patch. The conversation also includes tips on how to solve the problem, including reverse engineering the projectile problem to find the initial velocity and using the equation for horizontal motion.
  • #1
charliexx09
4
0
I'm really lost on how to do this physics problem because I don't really understand all the components of the problem:

A 7.25 kg block is sliding at 5.45m/s on a frictionless lab bench which is 8.6m high. The block hits a 1m "rough patch" before sliding off of the lab bench, and then lands 1.2m from the edge of the bench. What was the coefficient of friction between the block and the rough patch?

If you could show your work or explain how to do this it would be really helpful. Thank you! Also, this is my first post on the forums! :smile:
 
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  • #2
Ok, this is actually a really nice question! It provides a good mix between projectiles and Newton's laws.

Here's a tip reverse engineer the projectile problem to find the initial velocity of the projectile.
Then work on the part before the rough patch!
 
  • #3
How should I determine the initial velocity of the projectile if all that is given is is the distance in x & y and the acceleration? This leaves Time, Vi & Vf unknown?
 
  • #4
Here's a tip, Uy = 0 m/s

U = Viy
 
  • #5
Okay, here is my process so far:
I know that d= Vi*t + (0.5)(a)(t^2), so
-8.6= -4.9t^2
t= 1.325
Then, I plugged this into the equation for the horizontal
1.2=Vi(1.325) + (-4.9)(1.325)^2
So, Vi=7.398 m/s
This is where I am really stuck seeing as I was always bad at figuring out the coefficient of friction (I've only been doing Physics for a few months!). I know that Force of Fric= u*Fnormal ... is the force of friction here the same as the initial velocity? Thanks for all your help Darth Frodo :)
 
  • #6
Then, I plugged this into the equation for the horizontal1.2=Vi(1.325) + (-4.9)(1.325)^2

This is wrong I'm afraid, remember there is no acceleration in the x direction.

What the answer at the back of the book?
 

Related to 2D Kinematics Problem w/ Coefficient of Friction?

1. What is a coefficient of friction in 2D kinematics?

The coefficient of friction in 2D kinematics is a measure of the resistance between two surfaces in contact with each other. It is represented by the symbol "μ" and is used to calculate the force of friction between the two surfaces.

2. How do you calculate the coefficient of friction in a 2D kinematics problem?

To calculate the coefficient of friction, you need to divide the force of friction by the normal force between the two surfaces. The normal force is the perpendicular force exerted by one surface on the other.

3. What are some common units for coefficient of friction?

The coefficient of friction is a dimensionless quantity, meaning it has no units. However, it is commonly expressed in decimal form or as a percentage.

4. How does the coefficient of friction affect the motion of objects in a 2D kinematics problem?

The coefficient of friction plays a crucial role in determining the motion of objects in a 2D kinematics problem. It can either increase or decrease the force required to move an object, and can also affect the direction and speed of its motion.

5. Can the coefficient of friction be greater than 1 in a 2D kinematics problem?

Yes, the coefficient of friction can be greater than 1 in a 2D kinematics problem. This usually occurs when there is a high level of interaction between the two surfaces, such as in the case of rough or sticky surfaces.

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