An object with a mass of #5 kg# is on a surface with a kinetic friction coefficient of # 4 #. How much force is necessary to accelerate the object horizontally at # 32 m/s^2#?

1 Answer
Dec 25, 2015

Answer:

Let us use the #2^"nd"# Newton law in order to get this.

Explanation:

According to Newton #2^"nd"# law, the total force made on a body (i.e. the sum of all forces) is proportional to its acceleration, in this way:

#F_"Total"= sum F = m a#

In our question, there are two forces:
- Our own force, which we are going to call #F_"ours"#.
- The friction force, which we are going to call #F_"fr"#.

Both forces must be summed up, although with opposite sign, because friction always acts against movement.

We know that friction force can be obtained by:

#F_"fr" = mu cdot N#

with #mu# being the kinetic friction coefficient, and #N# being the normal force (equals to the weigth of the body in our problem).
So:

#F_"fr" = mu cdot N = mu cdot p = mu cdot (m g) = 4 cdot (5 "kg" cdot 9.8 "m/s"^2) = 196 "N"#

Now that we know this, we can rewrite #2^"nd"# Newton law as:

#sum F = F_"ours" - F_"fr" = m a rightarrow#
#rightarrow F_"ours" = ma + F_"fr" = 5 "kg" cdot 32 "m/s"^2 + 196 "N" = 356 "N"#

So, we must make a force of 356 N in order to move our object with an acceleration of #32 "m/s"^2#.