How Does Ramp Angle and Friction Affect Ice Block Speed?

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The discussion centers on calculating the angle of a frictionless ramp and the speed of an ice block sliding down it. The ice block, weighing 8.20 kg and released from a height of 1.43 m, reaches a speed of 2.11 m/s at the bottom. Participants suggest using the equations for acceleration down the ramp and normal force to solve for the angle and speed under friction. The introduction of a constant friction force of 11.0 N complicates the motion, requiring adjustments to the acceleration calculations. Overall, the focus is on applying physics equations to determine the effects of ramp angle and friction on the ice block's speed.
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A 8.20-kg block of ice, released from rest at the top of a 1.43-m-long frictionless ramp, slides downhill, reaching a speed of 2.11 m/s at the bottom.

What is the angle between the ramp and the horizontal?

What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 11.0 N parallel to the surface of the ramp?


I think I use the equations of ax=gsin(theta) and Fn=mgcos(theta)? I'm not sure what to do. Can someone please help me?

Thank you!
 
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With the acceleration component down the ramp, then find the distance traveled to achieve a velocity of 2.11 m/s. It starts at rest.

In the second part, the friction works against the weight component acting down the ramp, so the acceleration is less. Find that acceleration and solve as one does for the first part of the problem.
 
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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