Answer:
a)
b)
m = 48lb
c)
b = 144.76lb
Explanation:
The general equation of a damping oscillate motion is given by:
(1)
uo: initial position
m: mass of the block
b: damping coefficient
w: angular frequency
α: initial phase
a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:
Then, you obtain by replacing in (1):
6in = 0.499 ft
b.
mass, m = 48lb
c.
b = 144.76 lb/s
Answer:
Explanation:
Given that,
Mass of counterweight m= 4kg
Radius of spool cylinder
R = 8cm = 0.08m
Mass of spool
M = 2kg
The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:
Then we have,
τ(net) = R~ × T~
τ(net) = R~•i × mg•j
τ(net) = Rmg• k
τ(net) = 0.08 ×4 × 9.81
τ(net) = 3.139 Nm •k
The magnitude of the net torque is 3.139Nm
b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:
L~ = R~ × m~v + I~ω
L = mRv + MR v
L = (m + M)Rv
L = (4 + 2) × 0.08
L = 0.48 Kg.m
C. τ =dL/dt
mgR = (M + m)R dv/ dt
mgR = (M + m)R • a
a =mg/(m + M)
a =(4 × 9.81)/(4+2)
a = 6.54 m/s
Answer:
Before: 0 m/s
After: -4 m/s
Explanation:
Before: Since you and your beau started at rest, your beau initial velocity is 0 m/s.
After: Since we have to conserve momentum,
momentum before push = momentum after push.
The momentum before push = 0 (since you and your beau are at rest)
momentum after push = m₁v₁ + m₂v₂ were m₁ = your mass = 60 kg, v₁ = your velocity after push = 3 m/s, m₂ = beau's mass = 45 kg and v₂ = beau's velocity.
So, m₁v₁ + m₂v₂ = 0
m₁v₁ = -m₂v₂
v₂ = -m₁v₁/m₂ = -60 kg × 3 m/s ÷ 45 kg = -4 m/s
So beau moves with a velocity of 4 m/s in the opposite direction
V = m1 u1 - m2 u2 / m1
v = 0.01 * 500 - 2 * 1.4 / 0.01
v = 220 m/s
