Acceleration problem with friction

[Fuzbawl]

Homework Statement

A child is pushing a 20kg box across the floor. The coefficient of static friction between the box and the floor is 1.3. the coefficient of kinetic friction between the box and the floor is 1.0. If the child is pushing with a force of 400N, how far does the object travel in 20s?

Homework Equations

F=ma, Ffk=μk*Fn

The Attempt at a Solution

Since Fn (normal) is equal to the mass of the object in kg*9.8(g), I can substitute Newton's second law into the kinetic friction equation: Ffk=μk*F/a*g. Plugging in the given quantities from the problem I found : 400=1*400/a*9.8. However, when solving this, I find a=9.799m/s^2. I think that i did something wrong

 

[tiny-tim]

Welcome to PF!

Hi Fuzbawl! Welcome to PF!

(try using the X² and X₂ buttons just above the Reply box )

… Ffk=μk*F/a*g. Plugging in the given quantities from the problem I found : 400=1*400/a*9.8.

I'm confused …

there are two forces (the child and the friction), producing the acceleration.

 

[Fuzbawl]

There are indeed two forces working on the box. The child is pushing the box and friction is resisting. I know the box moves because the coefficiant of static friction is 1.3 Therefore the minimum amount of force required is = 254.8N.

so I have been working on it a bit more by subbing some formulas together and came up with m*a=Fa-μk*Fn where m is mass, a is forward acceleration, Fa is the applied force, μk is the coefficiant of kinetic friction and Fn is the normal force.

20*9.8=196

20a=400-1(196)
a=10.2m/s^2

 

[tiny-tim]

m*a=Fa-μk*Fn where m is mass, a is forward acceleration, Fa is the applied force, μk is the coefficiant of kinetic friction and Fn is the normal force.

20*9.8=196

20a=400-1(196)
a=10.2m/s^2

Looks good!

(but too many significant figures )

And now find the distance ​

 

[Fuzbawl]

On this particular assignment, the goal is to proge that I can solve the problem; I don't need to worry about significant figures, but thank you for reminding me. Finding the distance shouldn't be a problem although 10.2 m/s^2 seems quite large for a child pushing a box. It is for this reason that I posted here. Thanks for confirming what I wrote