Beginning at time t = 0, a student exerts a horizontal force on a box of mass 30 kg, causing it to move at 1.2 m/s toward an elevator door located 16 m away, as shown above. The coefficient of kinetic friction μk between the box and the floor is 0.20.
(b) Calculate the magnitude of the horizontal force the student must exert on the box in order to keep it moving at 1.2 m/s.

Question

Beginning at time t = 0, a student exerts a horizontal force on a box of mass 30 kg, causing it to move at 1.2 m/s toward an elevator door located 16 m away, as shown above. The  coefficient of kinetic friction μk between the box and the floor is 0.20.
(b) Calculate the magnitude of the horizontal force the student must exert on the box in order to keep it moving at 1.2 m/s.

anonymous · Answer

Hmm, I'm not sure if this is 100% correct, but I'm going to take a whack at it and someone else can correct me

First of all, let's draw a FBD:
|dw:1445404006993:dw|

We notice that the box is moving in the horizontal direction (x-direction). So let's see what else is happening here. 

The force that we applied is to the right and can be defined as $F_x=ma_x$

Next we need to look at the normal force; however, we only need it to find the frictional force (you'll see why later). The normal force is always the force opposite of the force into the surface. In this case, the force into the surface (floor) is just the weight, which is $-mg$. Therefore, $N=mg$

Next we have the frictional force, which is always defined as: $F_f=\mu N$ which for our situation means that $F_f=\mu mg$

Now let's sum what we have!
$$\sum F=F_x-F_f=ma_x- \mu mg$$
Now, here comes the tricky part.. If I am correct, the question says that the velocity remains constant (1.2m/s before and 1.2m/s after). What happens when velocity is constant? Position increases linearly, BUT acceleration is ZERO. This would mean that our term for $ma_x$is = 0 since a is 0. 

So $$\sum F=-F_f=- \mu mg$$
Plug in your numbers and you should get your answer!

Now here's the other trick... the FRICTIONAL force is acting in the leftward direction, which means that, according to our convention we made that right is positive, that the frictional force is negative. But what we're calculating here needs to be rightward (positive) because we need to keep pushing it at the same force as the force due to the kinetic friction. So I think the final answer should be positive but I'm not 100%

Perhaps @ganeshie8 or @Directrix can check my answer on this.