Minimum Speed of a Rock: Solving Problem with Pit Gap of 40m

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In summary, the block must have a minimum speed of 42.00m/s at the base of the hill to avoid falling into the pit.
  • #1
surfhare75
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Here is the Problem. A 2.8-kg block slides over the smooth, icy hill shown in the figure .The top of the hill is horizontal and 70 m higher than its base. i attached the Figure Below.


What minimum speed must the block have at the base of the hill so that it will pass over the pit at the far side of the hill without falling into it?

I tried using .5mv1^2+m1gH=.5mv2^2+m2gH but I need to know the gap between the pit which is 40m. I believe I use H=.5gt^2 to find the time and use projectile motion x=vosin(x)t to find the distance.
 

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  • #2
I can't see the picture.

Is the hill curved?
 
  • #3
The picture should be loaded now, but in the picture the hill is curved but does not seem to be stated specially. This is from a chapter on potential and conservation energy
 
  • #4
Cant' see it. It says: "Attachments Pending Approval"
 
  • #5
I also cannot see figure.:smile:
 
  • #6
I think if the top of the hill is horizontal,then we can use

[tex]x=v_{top}t[/tex]

[tex]y=\frac{1}{2}gt^2[/tex]
 
  • #7
The solution to this should be the initial velocity it takes to get to top of the hill plus the velocity needed to clear the pit, right? I get about 56.8 m/s which is apparently wrong.

[tex]\sqrt{1372}\;m/s+19.8\;m/s[/tex]
 
  • #8
Can someone please help ?
 
  • #9
I have found the answer.
 
  • #10
I'm not sure if this is right, but this is what I came up with:

1) I found the speed that the object needed to be going to clear the jump.. not too hard. I assumed the ground was horizontal at the launch angle.
2) I used Energy to solve the problem.

Speed to Clear:
[tex]\Delta[/tex]y = .5at2
-20m = -4.9[tex]m/s[/tex]2t2
t = 2.02 seconds

no wind resistance = no acceleration

v = delta x / t = 40m / 2.02s = 19.80m/s

--
Etop = Ebottom

2.8kg(g)(70m) + .5(2.8kg)(19.8[tex]m/s[/tex])2 = .5 (2.8kg)v2
v = 42.00[tex]m/s[/tex]
 
  • #11
Yes, that is correct.


[tex]\sqrt{1372\;m^2/s^2+(19.8\;m/s)^2} [/tex]= Life, the Universe, and Everything
 

FAQ: Minimum Speed of a Rock: Solving Problem with Pit Gap of 40m

What is the minimum speed required for a rock to clear a pit gap of 40m?

The minimum speed required for a rock to clear a pit gap of 40m can be calculated using the formula v = √(2gh), where v is the minimum speed, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the pit gap (40m). Therefore, the minimum speed required for a rock to clear a pit gap of 40m is approximately 28.2 m/s.

How does the weight of the rock affect its minimum speed?

The weight of the rock does not affect its minimum speed as long as it is being dropped from the same height. This is because the only factors that affect the minimum speed are the height of the pit gap and the acceleration due to gravity.

Can the minimum speed of a rock be greater than the calculated value?

Yes, the minimum speed of a rock can be greater than the calculated value. This is because the calculated value only takes into account the minimum speed required for the rock to clear the pit gap without hitting the bottom. In reality, the rock may need to have a higher velocity to account for air resistance and other external factors.

How does the angle of the pit gap affect the minimum speed of a rock?

The angle of the pit gap does not affect the minimum speed of a rock as long as the height of the pit gap remains the same. This is because the minimum speed is determined by the height of the pit gap and the acceleration due to gravity, not the angle of the gap.

What other factors can affect the minimum speed of a rock?

Other factors that can affect the minimum speed of a rock include air resistance, the shape and size of the rock, and the surface of the pit gap. These factors can alter the velocity of the rock and may require a higher speed to successfully clear the pit gap.

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