" 20 m/s² " means that if gravity is the ONLY force on the object
(the object is in 'free fall'), then its speed increases by 20 m/s
every second.
That's the answer to your question. Now, let me ask you
another one:
How does a speedometer tied to a falling rock work ?
How can it measure the rock's speed ?
Maybe one way would be to have a little tiny propeller on
the front of the speedometer, and it could measure how fast
the propeller is spinning as the rock falls through the air ?
Great idea. But we already said the rock is in free-fall,
so there's no air resistance, we can't have any air, and
there's nothing to spin the propeller.
How would you do it ? How can you measure the rock's speed ?
200N
Explanation:
600N-400N = 200N
Answer:
it is 30 newtons to the left
Explanation:
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
