Since you are given both the static and kinetic friction coefficients, this becomes two separate problems: First, how much force will it take to overcome the static friction and start the calculator moving from rest, and second, how much force will it then take to overcome kinetic friction and keep...

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Since you are given both the static and kinetic friction coefficients, this becomes two separate problems: First, how much force will it take to overcome the static friction and start the calculator moving from rest, and second, how much force will it then take to overcome kinetic friction and keep the calculator moving at a constant speed across the surface?

For step one we need the static friction equation

where **μ** = the coefficient of static friction and N = the normal force, which describes how hard the object is being pulled down by gravity.

(Note: in your question, 7.8N is the normal force; we can tell because it is given in Newtons. If you were to be given the mass of the object instead, you would express the mass in kilograms and then multiply that number times gravity (9.81m/s/s) to get the normal force in Newtons.)

So, to start the calculator sliding we must apply 0.5 x 7.8= 3.9 N of horizontal force.

To keep the calculator sliding, we do a similar calculation using

in which we use the coefficient of sliding friction.

So, to *keep* the calculator sliding once is has begun moving, you must apply 0.4 x 7.8N = 3.12 N of horizontal force.