Work-Energy Theorem: Calculating Work Done on a Sliding Block

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Homework Help Overview

The discussion revolves around the Work-Energy Theorem as applied to a block sliding on a horizontal surface. The original poster presents a problem involving calculating the work done to bring a block to rest, given its mass and initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between work and kinetic energy, questioning the application of the work-energy theorem. There is discussion about the calculation of work using kinetic energy and the implications of negative signs. Additionally, the impact of friction on acceleration is examined, with participants questioning whether friction is the only force acting on the block.

Discussion Status

The discussion is active, with participants providing guidance on the application of the work-energy theorem and clarifying the relationship between forces and acceleration. There is acknowledgment of the need to consider all forces acting on the block, and some participants are correcting their understanding of units.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. There is a focus on ensuring the correct application of physics principles without providing direct solutions.

kmikias
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I am just confuse in this question may be i forgot about work and energy .I just need little help not the answer ,here is the question

1.A block of mass 2.2 kg, which has an initial
velocity of 3.9 m/s at time t = 0, slides on a
horizontal surface.
Calculate the work that must be done on
the block to bring it to rest. Answer in units j.

solution .

Is it work = KE WHICH IS 1/2MV^2

I try to use W = F.d
but i don.t know the acceleration
 
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kmikias said:
Is it work = KE WHICH IS 1/2MV^2
Good. Use the work-energy theorem: W = ΔKE.

(What's the change in KE?)
 


oh thank you , now i got it, i didn't put negative sign
 


how about If a constant friction force of 11 Newtons is
exerted on the block by the surface, what is
the acceleration? Answer in units of m/s.

solution
F - friction force = mass * acceleration

is that right
 


kmikias said:
solution
F - friction force = mass * acceleration
Is friction the only horizontal force acting on the block? If so, then all you need is F = m*a, where F is the friction force.

Also: acceleration has units of m/s^2, not m/s (m/s is the units for speed)
 


thank you doc i just forgot to write the square
 

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