Coefficent of friction for a box moving at cosntant speed

AI Thread Summary
A horizontal force of 190N is required to pull a 70kg box at a constant speed, indicating that the force of kinetic friction (F_f) equals the applied force. Since the box moves at constant speed, the acceleration (a) is zero, meaning the net force is also zero, and the applied force balances the frictional force. The equation for kinetic friction is F_f = N * μ, where N is the normal force. Given that the normal force equals the weight of the box (N = mg), the coefficient of friction (μ) can be calculated using the known values. This analysis confirms that the frictional force is indeed 190N, allowing for the determination of the coefficient of friction.
jwxie
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Homework Statement


A horizontal force fo 190N is needed to pull a 70kg box across a horizontal floor at a constant speed. What is the coefficent of friction between the box and the floor?


Homework Equations



Fnet = ma

The Attempt at a Solution



My setup is F - F_f = m*a

But I have trouble getting the acceleration...
F_f is supposed to be F_f = N * u

I am guessing F_f is knifectic fricition? Is it 190N?

I am confused. Please help!
 
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jwxie said:

Homework Statement


A horizontal force fo 190N is needed to pull a 70kg box across a horizontal floor at a constant speed. What is the coefficent of friction between the box and the floor?


Homework Equations



Fnet = ma

The Attempt at a Solution



My setup is F - F_f = m*a

But I have trouble getting the acceleration...
F_f is supposed to be F_f = N * u

I am guessing F_f is knifectic fricition? Is it 190N?

I am confused. Please help!

Yes, Ff is the force of kinetic friction. Notice that the problem says that the box moves at a constant speed. So, what is the acceleration, then? Based on that, and Newton's Second Law (which you've written above), what must be true about how Ff compares to F (the applied force)?
 
a = 0? so it means they are the same force??

So it means all we need is to use the Ff equation now.
 
jwxie said:
a = 0? so it means they are the same force??

So it means all we need is to use the Ff equation now.

Yes, a = 0, so the forces must be the same in magnitude. (They have opposite directions, so they cancel each other out).

Yeah, so since you know the value of Ff, and you know N, you can solve for μ
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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