It takes a horizontal force of 375N to drag a box of mass 300kg at uniform speed along a level track. Calculate the coefficient of friction between the contact surfaces.

Question

asylum15 · Answer

mu = f/n?

anonymous · Answer

Friction force = N * mu
N = mg
Friction force = - 375 N

asylum15 · Answer

Archie didn't read that properly, lol.

Looking for mu :)

anonymous · Answer

But is that so difficult  to derive mu from given formulas?

anonymous · Answer

))

asylum15 · Answer

See my first message? I was wondering if I was correct ;)

If the box was then inclined at 40 degrees, the magnitude of the force parallel to the slope would be w.sin.theta?

anonymous · Answer

F = ma = 0. F - f = 0, so F = f, where f is frictional force. f = uN. 375N = u(300kg)(9.8m/s^2). u = .13

asylum15 · Answer

Any ideas for the 2nd part Aj? :)

anonymous · Answer

$$\mu $$=375$$\div $$3000 
hence $$\mu $$ = 0.125