Coefficient of friction problem

AI Thread Summary
The problem involves a 10 kg mass on a rough surface experiencing two different accelerations under varying forces. Initially, the mass accelerates at 5 m/s², and when the applied force is doubled, it accelerates at 18 m/s². The net force equations for both scenarios are set up to find the coefficient of friction. After some attempts and clarifications on the relationship between the forces, the correct coefficient of friction was determined. The discussion concludes with the participant successfully finding the answer.
kranav
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Homework Statement



a body of mass 10 kg lies on a rough horizontal surface with an acceleration of 5 m/s and when the horizontal force us doubled , it gets an acceleration of 18 m/s^2. Then the coefficient of friction between the body and the horizontal surface is:

A) 0.2 B) 0.8 C) 0.4 D) 0.6


Homework Equations


fnet = f - umg = ma1 1st case
Fnet = F - umg = ma2 2nd case

The Attempt at a Solution


I tried to solve it but u comes out to be zero.
 
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kranav said:
fnet = f - umg = ma1 1st case
Fnet = F - umg = ma2 2nd case
This looks good. (How do f and F relate?)

The Attempt at a Solution


I tried to solve it but u comes out to be zero.
Try it one more time.
 
Doc Al said:
This looks good. (How do f and F relate?)


Try it one more time.


thanks! got the answer!
 
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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