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#?

1 Answer
Feb 3, 2016

Answer:

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.)