Solve Coefficient of Kinetic Friction Problem: 6m/s, 25° Angle

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To solve the problem of a wooden block sliding down an inclined plane at a constant velocity of 6 m/s on a 25° angle, it's essential to understand that constant velocity indicates balanced forces acting on the block. The coefficient of kinetic friction can be determined by analyzing the forces involved, including gravitational force and frictional force. The angle of 25° plays a crucial role in calculating the components of these forces. A force diagram is recommended to visualize the forces acting on the block, which will aid in deriving the necessary equations. Overall, the discussion emphasizes the importance of understanding the relationship between velocity, angle, and friction in solving the problem.
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I can not figure out the formula for the following question about friction. Need some help! Awooden block slides directly down an inclined plan at a constant velocity of 6m/s, how large is the coefficient of the kenetic friction, if the plane makesan angle of 25 degrees with the horizontal?
 
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Have you drawn a force diagram for the block?

What does constant velocity tell you about the forces?

Where does the 25 degrees come into the calculation? How about the friction coefficient?

Can you show us what you've done so far?
 
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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