A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.
1 answer:
Answer:
48.7 J
Explanation:
For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.
In particular:
- The elastic potential energy is maximum when the system is at its maximum displacement
- The kinetic energy is maximum when the system passes through the equilibrium position
Therefore, the maximum kinetic energy of the system is given by:
where
m is the mass
v is the speed at equilibrium position
In this problem:
m = 3.6 kg
v = 5.2 m/s
Therefore, the maximum kinetic energy is:
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we know that center of mass is given as
r = (m₁ + m₂
)/(m₁ + m₂)
taking derivative both side relative to "t"
dr/dt = (m₁ d/dt + m₂ d
/dt)/(m₁ + m₂)
v = (m₁ + m₂
)/(m₁ + m₂)
taking derivative again relative to "t" both side
dv/dt = (m₁ d/dt + m₂ d
/dt)/(m₁ + m₂)
a= (m₁ + m₂
)/(m₁ + m₂)
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d= * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip