Help with Part C: Horizontal Force for Moving Two Trunks

[phy_]

help with part c??

A worker of a moving company places a 252 kg trunk on a piece of carpeting and slides it across the floor at constant velocity by exerting a horizontal force of 425 N on the trunk.

a) what is the coeffient of kinetic friction

Fn = mg
= 252 kg x 9.8 N/kg
= 2469.6 N

kinetic coefficient equals 425 N/2469.6 N = 0.17

b) what happens to the coefficient of kinetic friction if another 56 kg trunk is splaced on top of the 252 kg trunk?

Fn = mg
=(308 kg) (9.8 N/kg)
=3018.4

kinetic coefficient equals 425 N/3018.4 N = 0.14

c) what horizontal force must the mover apply to move the combination of the two trunks at constant velocity

the answer is 5.2 x 10^2 N any help would be appreciated as to how to solve c
thank-you

 

[Sumanyu]

what horizontal force must the mover apply to move the combination of the two trunks at constant velocity


force is not being applied to an object moving an constant velocity or sitting "still"...

force = mass x acceleration...

I don't see how you'd need a force to move an object at a constant velocity

 

[The Bob]

You need a force because, although moving at constant velocity, there is a coefficent of friction that has to be overcome.

Also, please don't post a question twice.

The Bob

 

[LowlyPion]

In b) can you explain how the coefficient of friction would change with more weight?

 

[phy_]

i am not sure other than i added 56 kg (i had made an error and written 5 earlier)

 

[phy_]

is b correct?

 

[LowlyPion]

i am not sure other than i added 56 kg (i had made an error and written 5 earlier)

If it has a coefficient of friction from a), what is the coefficient of friction?

Moving force/Normal force ?

Isn't it constant? If you change the normal force doesn't the moving force need to be bigger?

Isn't that what c) is about?