Answer:
Explanation:
General Equation of SHM is given by
where x=position of particle
A=maximum Amplitude
angular frequency
t=time
At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e.
where k=spring constant
Potential Energy is given by
also it is given that Potential Energy(U) is equal to Kinetic Energy(K)
Total Energy
Total
at
velocity is
Answer:
Explanation:
We shall take the help of vector form of displacement . Taking east as i and north as j
4.0m N = 4 j
7.5 m E = 7.5 i
6.8 m S = - 6.8 j
3.7 m E, = 3.7 i
3.6 m S = - 3.6 j
5.3 m W = - 5.3 i
3.7 m N, = 3.7 j
5.6 m W = - 5.6 i
4.4 m S = - 4.4 j
4.9 m W = - 4.9 i
Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i
= -4.6 i -7.1 j
magnitude of displacement =
= 8.46 m
Direction
Tanθ = 7.1/ 4.6
θ = 57⁰ south of west .
distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9
= 49.5 m
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s