A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional force opposing the motion?

In order to solve this problem, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

[tex]\sum F = ma[/tex]

where

[tex]\sum F[/tex] is the net force

m is the mass

a is the acceleration

Here we are just interested in the situation along the horizontal direction. The net force acting on the block is:

[tex]\sum F = F_a -F_f[/tex]

where

[tex]F_a = 40 N[/tex] is the applied force (forward)

[tex]F_f[/tex] is the frictional force (backward)

m = 5 kg is the mass of the block

[tex]a=5 m/s^2[/tex] is the acceleration

Substituting into the equation and solving, we find the frictional force:

[tex]F_a-F_f = ma\\F_f = F_a-ma=40-(5)(5)=15 N[/tex]

Learn more about Newton's second law:

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