Fnet=P-Pc=24-117.6=-93.6NFind Force Needed to Move 4kg Box in 3s

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SUMMARY

The discussion focuses on calculating the force required to move a 4kg box from rest to a velocity of 6m/s in 3 seconds, considering a coefficient of friction of 0.05. The acceleration is determined to be 2m/s², leading to a gravitational force of 39.2N. The net force is calculated as 117.6N, and the applied force necessary to overcome friction and achieve the desired acceleration is derived from these values.

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Homework Statement


A certain force moves a 4kg box from rest to a velocity of 6m/s in 3 seconds. If the coefficient of friction between the box and floor is 0.05, what is the magnitude of the force?

m=4kg
v0=0
v=6m/s
t=3s
μk=0.05

Homework Equations


a=v/t
F0=mg
P=Ft
P=mv

The Attempt at a Solution


a=(6)/(3)=2m/s
F0=(4)(9.8)=39.2N
Pc=(39.2)(3)=117.6
P=(4)(6)=24
 
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bearsa113609 said:

Homework Statement


A certain force moves a 4kg box from rest to a velocity of 6m/s in 3 seconds. If the coefficient of friction between the box and floor is 0.05, what is the magnitude of the force?

m=4kg
v0=0
v=6m/s
t=3s
μk=0.05

Homework Equations


a=v/t
F0=mg
P=Ft
P=mv

The Attempt at a Solution


a=(6)/(3)=2m/s
F0=(4)(9.8)=39.2N
Pc=(39.2)(3)=117.6
P=(4)(6)=24

HINT:
You've found the acceleration. Can you use that to find the net force, and then, from there, the unknown applied force?You don't need momentum or impulse for this problem.
 


thank you so much
 

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