Calculating Coefficent of Kinetic Friction: Box on Floor

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To calculate the coefficient of kinetic friction for a box weighing 325 N pushed with a 425 N force at a 35.2-degree angle, it's essential to analyze the forces acting on the box. A free body diagram should be drawn to visualize these forces, ensuring the net force is zero since the box moves at constant velocity. The forces must be resolved into horizontal and vertical components, leading to two equations: one for horizontal motion and one for vertical equilibrium. The frictional force can be expressed as μN, where μ is the coefficient of friction and N is the normal force. Solving these equations simultaneously will yield the value of μ.
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Help on calculating Coefficent of kinetic friction

Here is the problem: A box of books weighing 325 N moves with a constant
velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Find the Coefficent of kinetic friction between the box and the floor.

Thanks for the help!
 
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1. Draw a vector free body diagram showing all forces acting on the box.
2. Since it is moving at constant velocity the net force must be zero.
3. Resolve forces horizontally and vertically to get two equations. Remember that the friction force is \mu N, where \mu is the coefficient of friction and N is the normal reaction force.
4. Solve your two equations simultaneously to find \mu.
 
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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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