# 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?

Feb 3, 2016

There are two components of the required force: one to overcome the frictional force and one to accelerate the object. The total force is $F = m a + \mu m g = 2 \cdot 3 + 9 \cdot 2 \cdot 9.8 = 6 + 176.4 = 182.4 N$

#### Explanation:

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

$F = m a$

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

${F}_{\text{frict" = muF_"norm}}$ where $\mu$ is the frictional coefficient and ${F}_{\text{norm}}$ is the normal force.

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

${F}_{\text{norm}} = m g$

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

$F = m a + {F}_{\text{frict" = ma + muF_"norm}} = m a + \mu m g$

Substituting in the values from the question:

$F = m a + \mu m g = 2 \cdot 3 + 9 \cdot 2 \cdot 9.8 = 6 + 176.4 = 182.4 N$

(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.)