What horizontal force is required to accelerate the block

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Homework Help Overview

The problem involves determining the horizontal force required to accelerate a block of mass 4.0 kg at 5.0 m/s² while accounting for kinetic friction with a coefficient of 0.25.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of free body diagrams and relevant equations. They explore the relationship between net force, frictional force, and the applied force required for acceleration.

Discussion Status

Participants have provided various equations and calculations, questioning the setup of forces involved. There is an ongoing exploration of how to correctly account for friction and net force, with some guidance offered on the relationships between these forces.

Contextual Notes

There is a focus on ensuring that vertical and horizontal forces are treated appropriately in the equations. Participants are also clarifying the distinction between net force and total applied force in the context of the problem.

a7med2009
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A block of mass 4.0kg rests on a horizontal surface. What horizontal force is required to accelerate the block at 5.0m/s2 if the coefficient of kinetic friction is 0.25?
 
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start with writing a free body diagram.
& any equations you may know that you think would be useful here.
 


ΣF=ma

Fk=mu * N
 


ma=4.0kg*5.0m/s^2
=20 N
Fk= 0.25 * 20
= 5 NIs it correct?
 


so, how much force do you need to apply in order to overcome the frictional force?
 


ΣF=F-fk=mg
F=mg-fk
=(4.0)*(9.8)-(0.25)(39.2)
=29.4 N
 


In your last post, you've got a vertical force (mg) and a horizontal force (fk) in the same equation. Not allowed.

In your previous post, you found the frictional force's maximum value. You also found that in order to accelerate at 5m/s/s, you need 20 Newtons of NET force.

So how much TOTAL force must be applied (horizontally) to overcome friction and give the block enough force to accelerate?
 
Last edited:


ohh 0_o

it's
ΣF=F-fk=ma
F=ma+fk
=(4.0)*(5)+(0.25)(39.2)
=20+9.8
=29.8 N
 


looks correct to me
 

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