A 2.8-kg block slides over the smooth icy hill -- Find the minimum speed to clear the pit

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To determine the minimum speed for a 2.8-kg block to clear a pit after sliding over an icy hill, the horizontal distance of 160 meters must be considered. The discussion highlights the need to identify the horizontal force acting on the block, as this force is crucial for calculating work done. The relationship between work (W), force (F), and distance (d) is emphasized, suggesting that the force must be horizontal to apply the work-energy principle effectively. Participants inquire about the specific horizontal force that influences the block's motion. Understanding these dynamics is essential for solving the problem accurately.
nouvelague
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Homework Statement
A 2.8-kg
block slides over the smooth, icy hill shown in the figure (Figure 1). The top of the hill is horizontal and 70 m
higher than its base.
Figure1 of 1

Part A
What minimum speed must the block have at the base of the hill so that it will not fall into the pit on the far side of the hill?
Express your answer using two significant figures.

I'm assuming that since this has to go over the pit, it would have to then travel 160m and since it is horizontal distance it would have something to do with Fx, and therefore W=Fd? But also, there is going to be a vertical component, in that case mgh as well, and I have to solve for speed, so I need kinetic energy as well. But am I am not sure how to approach these problems or arranging the equations.
Relevant Equations
W=Fd
KE = 1/2 mv^2
PE = mgh
Screenshot 2024-10-19 211539.png
 
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nouvelague said:
it would have to then travel 160m and since it is horizontal distance it would have something to do with Fx, and therefore W=Fd?
What would F be here? If you are going to multiply it by a horizontal distance then it must be a horizontal force. What horizontal force?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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