A 1.75 kg box is pushed with a 8.35 N force across ground where k = 0.267. What is the net force on the box?

The net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.

To find the answer, we have to know more about the basic forces acting on a body.

How to find the net force on the box?

Let us draw the free body diagram of the given box with the data's given in the question.
From the diagram, we get,

[tex]N=mg\\F_t=ma\\F_t=F-f[/tex]

where, N is the normal reaction, mg is the weight of the box, [tex]F_t[/tex] is the net force, f is the kinetic friction.

We have the expression for kinetic friction as,

[tex]f=kN=kmg=0.267*1.75*9.8= 4.58N[/tex]

Thus, the net force will be,

[tex]F_t= 8.35-4.58=3.77N[/tex]

Thus, we can conclude that, the net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.

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aadildhillon023 aadildhillon023 · Answer 2 · 2022-07-30T14:28:37+02:00

The net force acting on the box is 3.77 N.

We need to learn more about the fundamental forces affecting a body in order to locate the solution.

How can I determine the box's net force?

Let's use the information provided in the question to draw the free body diagram of the given box.

The diagram gives us,

[tex]N=mg\\F_n=F-f[/tex]

where N stands for the normal reaction, mg for the box's weight, is the net force, and f for kinetic friction.

The kinetic friction expression is as follows:

[tex]f=kN=kmg=4.58N[/tex]

the net force will be as follows:

[tex]F_n=3.77N[/tex]

Thus, it is clear that the box with a mass of 1.75 kg will experience a net force of 3.77 N when it is pushed across a surface with a coefficient of friction of 0.264.

Learn more here about the fundamental forces here:

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