Calculating Initial Push Force for Box on Floor | Friction Problem

[zorro]

Homework Statement

A mover pushes a box weighing 200 N along a floor. The coefficient of static friction between the box and the floor is 0.8, while the coefficient of kinetic friction between the box and the floor is 0.2. The mover pushes the box from rest to a constant speed of 10 m/s.

How much harder must the mover push initially to get the box going than to maintain the box’s speed at 10 m/s?

The Attempt at a Solution

Force due to static friction = 200*0.8 = 160 N which should be the answer as the force is applied 'initially'. The answer given is 120 N which ridicules me.

 

[rock.freak667]

At 10 m/s, what is the force due to kinetic friction?

The difference between static and kinetic will give you "How much harder must the mover push initially".

 

[zorro]

Static frictional force acts till the object starts moving. It's maximum value is 160N. If you apply 160-40 = 120 N to the block, it won't move at all.

 

[bmxicle]

That's true that if you apply only 120N to the block it will not start to move, but the question is asking what the difference in force required to start the block moving vs the force required to push the block once it is moving.

 

[zorro]

hmm... I guess I did not understand the question earlier.