mjakstaite
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#1
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"The miners working in a salt mine use smooth wooden slides to move quickly from one level to another
Vertical drop = 15m
Mass = 90Kg
B) Calculate the maximum possible speed that the miner could reach at the bottom of the slide. "
I honestly have no idea on how to do tihs. I was able to do part A relatively easy and found PGE to be 13500J however we haven't been taught how to find maximum speed.
HELP i beg
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username3442196
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#2
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Equate the GPE to KE (1/2 m v^2)
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8013
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mgh=1/2mv^2
13500=1/2*90*v^2
v^2=13500/45=300
v=17.3 (3.s.f)
Please thank me later
Last edited by 8013; 2 years ago
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JaredzzC
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Conservation of energy. The energy you worked out for your Gravitational Potential has been converted to Kinetic energy.

At the bottom of the slide your potential energy is 0 because it has all been convereted to kinetic energy, so you equate Gravitational Potential to Kinetic Energy.

 mgh = \frac{1}{2} mv^2

You already worked out gravitational potential to be 13500J therefore your equation becomes  13500 = \frac{1}{2} mv^2 and you can rearrange for v.

Remember, energy can never be created or destroyed so it is converted into another form of energy.
Last edited by JaredzzC; 2 years ago
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JaredzzC
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(Original post by 8013)
mgh=1/2mv^2
13500=1/2*90*v^2
v^2=13500/45=300
v=17.3 (3.s.f)
Please thank me later
Appreciate you helping the OP but posting full solutions are against the forum guidelines.
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mjakstaite
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(Original post by JaredzzC)Conservation of energy. The energy you worked out for your Gravitational Potential has been converted to Kinetic energy.

At the bottom of the slide your potential energy is 0 because it has all been convereted to kinetic energy, so you equate Gravitational Potential to Kinetic Energy.

 mgh = \frac{1}{2} mv^2

You already worked out gravitational potential to be 13500J therefore your equation becomes  13500 = \frac{1}{2} mv^2 and you can rearrange for v.

Remember, energy can never be created or destroyed so it is converted into another form of energy.

Thanks i couldn't for the life of me find the formula. It was too late for physics homework clearly :')
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Iqra Haq
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oh my god i have the same exact example! thanks for this thread! do you go to parrs wood high school by any chance?
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ChungusUp
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(Original post by Iqra Haq)
oh my god i have the same exact example! thanks for this thread! do you go to parrs wood high school by any chance?
what year
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Iqra Haq
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(Original post by ChungusUp)
what year
year 10 what about you?
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ChungusUp
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(Original post by Iqra Haq)
year 10 what about you?
nah im not in parrswood but do you know a guy in yr 9 called aryas or a guy in yr11 called bobi?
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Iqra Haq
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(Original post by ChungusUp)
nah im not in parrswood but do you know a guy in yr 9 called aryas or a guy in yr11 called bobi?
oh sorry no i dont
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Har2005
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#12
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Holy crap I have the exact same 3 questions about the miner
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Laura606
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What was the gravitational field strength as I have the same question but not sure if done the answer working out wrong ?
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OmegaJJ
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K.E. = 1/2 m v2So you do √(13500 ÷ (0.5 × 90))and the answer is 17.3205080757 and I rounded it to 3 significant figures so it would be 17.3 m/s.
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