"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>
<span>This is an agrarian society, taken to its extreme. These societies are largely dependent upon farming and related activities as a way of earning income, and also for using the farmed items as a way of supporting oneself as food and clothing.</span>
Answer:
k = 49 N/m
Explanation:
Given that,
Mass, m = 250 g = 0.25 kg
When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.
The force due to mass is balanced by its weight as follows :
mg=kx
So, the spring constant of the spring is 49 N/m.
To determine the amount in grams of the iron, we need data on the density of iron. From literature, it has a value of <span>p=7.9 g/cm3. We simply multiply the volume to the density. We do as follows:
mass = 3.70 (7.9) = 29.23 g Fe
Hope this answers the question. Have a nice day.</span>
Is there information in the previous question which relates to this one?
