Physics Puzzler: Solving a Grade 12 Cliff Pushing Problem

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The discussion centers on a Grade 12 physics problem involving a box being pushed off a cliff. The box weighs 10.8 kg and is pushed with a constant horizontal force until it falls from a height of 12.96 m, landing 7.5 m from the base of the cliff. Friction acting on the box is 49.0 N, and the user is struggling to calculate the force applied by Bob. Despite attempts to solve the problem, the user is confused about the next steps after calculating a net force of 117.504 N. The thread highlights the challenges students face in applying physics concepts to real-world scenarios.
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Physics puzzler! Need serious help!

HELP! I am in grade 12 physics and my teacher gave us a puzzler for the weekend. I need serious help with it and all of my classmates can't figure it out either! The question is :

A 10.8 kg box is 5.0 m from the edge of a 12.96 m high cliff. Bob pushes the box over the edge of the cliff by applying a constant horizontal force to it until the box goes over the edge. The force of friction on the box is 49.0 N and the box lands 7.5 m from the base of the cliff. How much force did Bob apply to the box while he was pushing?

Please help me if you can. How do I go about even starting the problem?
 
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I did what the first post said to do and I still can't get it. I end up with Fnet= 117.504 N. What the hell am I supposed to do? I've spent hours on this problem and it's driving me crazy!
 
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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