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Answer :
The frictional force is 39.4 N
Explanation:
We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write
[tex]\sum F = ma[/tex]
where
[tex]\sum F[/tex] is the net force
m is the mass
a is the acceleration
Here we know that the box is moving with constant velocity, so its acceleration is zero:
[tex]a=0[/tex]
This means that the net force is also zero:
[tex]\sum F=0[/tex]
The net force on the block is given by the applied force, forward, and the frictional force, backward:
[tex]\sum F = F_a-F_f=0[/tex]
where
[tex]F_a=39.4 N[/tex] is the applied force
[tex]F_f[/tex] is the frictional force
Therefore, solving for [tex]F_f[/tex],
[tex]F_f=F_a=39.4 N[/tex]
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