What is the speed of a block sliding on a plank when it comes to rest?

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Homework Help Overview

The discussion revolves around a block sliding on a horizontal plank, subject to a frictional force that is 35% of its weight. The block comes to rest from an initial speed represented by the variable x. Participants are exploring how to determine the value of x based on the given conditions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express uncertainty about the problem's setup, particularly regarding the representation of x as speed or distance. There are discussions about the forces acting on the block, including gravitational and normal forces, and how these relate to the frictional force. Some participants attempt to derive acceleration and relate it to the kinematic equations.

Discussion Status

The discussion is progressing with participants clarifying the forces involved and how they relate to the acceleration of the block. There is an ongoing exploration of kinematic equations and the implications of the frictional force on the block's motion. While some participants have made attempts to solve for variables, there is no explicit consensus on the final approach yet.

Contextual Notes

Participants note that the problem lacks specific numerical values and relies on variable representation. The time factor is also introduced later in the discussion, indicating potential constraints in the initial problem statement.

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Homework Statement


A block, moving along a horizontal plank, is acted upon by a force of friction that is equal to 35% of the weight of the block. The block comes to rest from a speed of x m/s. Find x.

Homework Equations


v=u + at
F=F1 + F2

The Attempt at a Solution



I can't even attempt this because I have no clue where to start. They give you basically no information
 
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physics(L)10 said:

Homework Statement


A block, moving along a horizontal plank, is acted upon by a force that is equal to 35% of the weight of the block. The block comes to rest from a speed of x m/s. Find x.


Homework Equations


v=u + at
F=F1 + F2



The Attempt at a Solution



I can't even attempt this because I have no clue where to start. They give you basically no information

They give you information in the form of variables. They want an answer in the form of variables. Just to make sure, are they using x to represent speed, or x to represent distance traveled horizontally? This makes a difference.
 
They are using x to represent speed

Attempt (most likely wrong):

Equation to use: F=ma

You can find a relative mass (35% of the block weight) that stops the block. So that will =m. Force I don't know and a I don't know. I'm guessing you find a relative value of a which then can be used to find the speed.
 
physics(L)10 said:
They are using x to represent speed

Attempt (most likely wrong):

Equation to use: F=ma

You can find a relative mass (35% of the block weight) that stops the block. So that will =m. Force I don't know and a I don't know. I'm guessing you find a relative value of a which then can be used to find the speed.

Well you have two forces on the mass in the vertical, do you know what these are? They must be equal in magnitude and opposite in direction since the mass is not accelerating in the vertical.
And one force acting in the horizontal, friction. They are telling you that the frictional force is a portion of one of the forces in the vertical.

Here would be a picture of the forces assuming the mass is moving to the right. Do you know what the different forces are called. If so we can figure out a and get this done with some easy kinematics.

forces.JPG
 
The down force is the block pushing against the floor and the up force is the ground pushing back against the block (normal force), correct?
 
physics(L)10 said:
The down force is the block pushing against the floor and the up force is the ground pushing back against the block (normal force), correct?

Almost, the down force is due to gravity, mg... the Earth pulling on the mass... and yes, the ground supplies the normal force back on the block. So now what can we say about the frictional force in terms of mg?
 
35%mg?
 
physics(L)10 said:
35%mg?

yep or 0.35mg which is equal to the frictioanl force which you already called ma. So now solve for a...

Once you got a, do you have kinematic equations to help you answer the question?
 
v=u + at

where v=final velocity=0, u=initial velocity, a=acceleration=.35mg, t=time=20 seconds

I forgot to add the time into the original question. My bad.
 
  • #10
physics(L)10 said:
v=u + at

where v=final velocity=0, u=initial velocity, a=acceleration=.35mg, t=time=20 seconds

I forgot to add the time into the original question. My bad.

hold on...

0.35mg = ma so the m's cancel right? so after this you got a, v, t, and you are trying to determine u, which for some reason I can't figure is being called x? So you got everything you need.
 
  • #11
So (.35)(9.8)=a, v=0, t=20 and then we get u :)
 
  • #12
physics(L)10 said:
So (.35)(9.8)=a, v=0, t=20 and then we get u :)

yep...

so apparently friction is causing your mass to accelerate in the opposite direction it is moving bringing it to a halt v =0 over a 20 sec time period.
 
  • #13
Yuppp I get it :d thank sooooo much
 

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