Elastic potential energy.
When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.
Answer:
If the final question is; at what velocity will the first block start to move outward in m/s?
Explanation:
The motion have the velocity that will make the block move using:
μ
Resolving:
Answer:
38 N, 40.0° below the horizontal
Explanation:
Force exerted by an object equals mass times acceleration of that object: F = m ⨉ a. To use this formula, you need to use SI units: Newtons for force, kilograms for mass, and meters per second squared for acceleration.
Answer:
Magnitude = 3.64 × 
စ = 43.9°
Explanation:
given data
ship to travel = 1.7 × kilometers
turn = 70°
travel an additional = 2.7 × kilometers
solution
we will consider here
Px = 1.7 ×
Py = 0
Qx =2.7 × cos(70)
Qy= 2.7 × sin(70)
so that
Hx = Px + Qx ............1
Hx = 2.62 ×
and
Hy = Py + Qy ..........2
Hy = 2.53 ×
so Magnitude = 
Magnitude = 3.64 ×
so direction will be
tan စ = Hy ÷ Hx ......................3
tan စ = 
tan စ = 0.9656
စ = 43.9°
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
