Question

An object with a mass of `2 kg` is on a surface with a kinetic friction coefficient of `9`. How much force is necessary to accelerate the object horizontally at `3 ms^-2`?

Answer 1

There are two components of the required force: one to overcome the frictional force and one to accelerate the object. The total force is `F = ma + mumg = 2*3+9*2*9.8= 6 +176.4 = 182.4 N`

Explanation:

To accelerate a mass on a frictionless surface, it is just necessary to apply the force described by Newton's Second Law:

`F=ma`

In this case, an additional force is required to overcome the frictional force, which is defined this way:

`F_"frict" = muF_"norm"` where `mu` is the frictional coefficient and `F_"norm"` is the normal force.

The normal force in this case is just the weight force of the object:

`F_"norm"=mg`

Pulling it all together to find the total force, we get:

`F=ma + F_"frict" = ma + muF_"norm" = ma + mumg`

Substituting in the values from the question:

`F = ma + mumg = 2*3+9*2*9.8= 6 +176.4 = 182.4 N`

(I remarked on another similar question, but it bears repeating here: 9 is not a reasonable or sensible value for a coefficient of friction. The fault is with the teacher or book asking the question in the first place, not the student asking it here. Friction coefficients are typically between 0 and 1, sometimes slightly above 1, in some extreme cases approaching 2, but 9 is just way off the scale. 0.9 would be a much more sensible number.)