To accelerate a 4.0 kg block at 5.0 m/s² on a surface with a coefficient of kinetic friction of 0.25, a total horizontal force must be calculated. The net force required for acceleration is 20 N, derived from the equation ΣF = ma. The frictional force, calculated as Fk = μ * N, equals 9.8 N. Therefore, the total force needed to overcome friction and achieve the desired acceleration is 29.8 N. This calculation confirms the necessary force to move the block effectively.
#1
a7med2009
8
0
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?
start with writing a free body diagram.
& any equations you may know that you think would be useful here.
#3
a7med2009
8
0
ΣF=ma
Fk=mu * N
#4
a7med2009
8
0
ma=4.0kg*5.0m/s^2
=20 N
Fk= 0.25 * 20
= 5 NIs it correct?
#5
graphene
195
2
so, how much force do you need to apply in order to overcome the frictional force?
#6
a7med2009
8
0
ΣF=F-fk=mg
F=mg-fk
=(4.0)*(9.8)-(0.25)(39.2)
=29.4 N
#7
merryjman
183
0
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?
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#8
a7med2009
8
0
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