Find minimum speed of a block so that it makes it up over a hill and jump a pit? : Eco Answers

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Question by Miximinix: Find minimum speed of a block so that it makes it up over a hill and jump a pit?
The block’s mass is 2.8 kg. The top of the hill is 70 m high and located 120 m from the starting point. The hill is not a straight line or straight curve, it has a small bump in it . After the top of the hill, there is a deep pit 40 m wide. We are trying to get the block over this pit to the other side, which is 50 m high. What is the minimum speed that the block needs to make it over the pit?

This is part of the “potential energy and energy conservation” chapter. I’m going nuts trying to figure this out, thanks in advance for any and all help!

Best answer:

Answer by Harald M
Here’s my attempt:

We need to make it up to 70m; and have enough speed there to jump over that 40m wide pit so that we land on the other rim, which is 20m lower. I do not see what for we’d need the 120 m …
We completely ignore air resistance and friction – nothing said in the problem statement about this.

We compute it from the “back”: Approximating standard gravitational acceleration with 10m/s², how long does it take the block to fall 20m? Well, if that time is t seconds, then the falling speed at the end will be 10t m/s (v = a*t). The average speed over that time will have been 5t m/s, and hence the distance fallen will have been 5t*t m. If this is equal to 20m, then t = 2s.
So we have 2 s to jump over that pit; which means we need a speed of 40 m / 2s = 20m/s at the top of the hill.

The kinetic energy, up there, must therefore be at least mv²/2 = 2.8*20²/2 J = 200*2.8 J = 560 J.

The potential energy consumed by running up that 70m is mgh = 2.8*10*70 J = 1960 J.

Together, we therefore need a “start energy” of 1960 J + 560 J = 2520 J.

We supply this as kinetc energy with speed u, so that we have

2520 = 2.8*u²/2
5040 = 2.8*u²
1800 = u²
u = 42.42 m/s

Hope that helped (but please check my computations or, better, redo them)!

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