# An object with a mass of #6 "kg"# is on a surface with a kinetic friction coefficient of #15#. How much force is necessary to accelerate the object horizontally at #1 "m/s"^2#?

##### 1 Answer

Mar 3, 2016

#### Answer:

#### Explanation:

From Newton's Second Law, we know that

#Sigma vecF = m * veca#

Now, to know

The weight is balanced with the normal force. All that is left is the applied force and the kinetic friction. The applied force is stronger than the kinetic friction, or else the object would not speed up.

So we write

#F_"applied" - F_"friction" = m * a#

To calculate

#F_"friction" = (15) * (6 "kg") * (9.8 "m/s"^2)#

#= 882 "N"#

Now, substitute all the values we have.

#F_"applied" - 882 "N" = (6 "kg") * (1 "m/s"^2)#

#F_"applied" = 6 "N" + 882 "N"#

#= 888 "N"#

#~~ 890 "N"#