Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = ![\frac{1}{3}\pi r^2 h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20h)
thus,
volume of the cone =
= 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the
times the total height
thus,
center of mass lies at, h' = ![\frac{1}{4}\times4=1m](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%5Ctimes4%3D1m)
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
Answer:
![T= 2\pi\times sqrt(m/(k1+k2))](https://tex.z-dn.net/?f=T%3D%202%5Cpi%5Ctimes%20sqrt%28m%2F%28k1%2Bk2%29%29)
Explanation:
When the block is displaced by x units
F= spring force
two springs are connected parallel
![F =-k_1x - k_2x](https://tex.z-dn.net/?f=F%20%3D-k_1x%20-%20k_2x)
Writing Newtons second law, F = ma
![-k_1x - k_2x =ma](https://tex.z-dn.net/?f=-k_1x%20-%20k_2x%20%3Dma)
![-k_1x - k_2x = mx''](https://tex.z-dn.net/?f=-k_1x%20-%20k_2x%20%3D%20mx%27%27)
a= x" ( differentiating x w.r.t time twice)
![x''+(k_1/m + k_2/m) x=0](https://tex.z-dn.net/?f=x%27%27%2B%28k_1%2Fm%20%2B%20k_2%2Fm%29%20x%3D0)
this the standard form of equation of oscillation spring mass system
This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration
since, Period ![T=2\pi\sqrt{\frac{m}{K_{eq.}} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%5Csqrt%7B%5Cfrac%7Bm%7D%7BK_%7Beq.%7D%7D%20%7D)
therefore,
![T= 2\pi\times sqrt(m/(k1+k2))](https://tex.z-dn.net/?f=T%3D%202%5Cpi%5Ctimes%20sqrt%28m%2F%28k1%2Bk2%29%29)
Alkali Metals ......................................