Question

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?

Answer 1

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,

                                 $N=mg\\F_t=ma\\F_t=F-f$

where, N is the normal reaction, mg is the weight of the box, $F_t$ is the net force, f is the kinetic friction.

• We have the expression for kinetic friction as,

                      $f=kN=kmg=0.267*1.75*9.8= 4.58N$

• Thus, the net force will be,

                        $F_t= 8.35-4.58=3.77N$

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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Answer 2

When a box with a mass of 1.75 kg is pushed across a surface with a coefficient of friction of k=0.267 and a force of 8.35 N, 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,

                             $N=mg\\F_n=F-f$

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:

                            $f=kN=kmg=4.58N$

• the net force will be as follows:

                                     $F_n=3.77N$

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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