A 0.6 kg block is on a frictionless horizontal surface and is attached to a ligh
ID: 1470216 • Letter: A
Question
A 0.6 kg block is on a frictionless horizontal surface and is attached to a light spring. A horizontal force of 4.25 N is required to hold the block at rest when pulled 5cm from its equilibrium position. The block is released and undergo simple harmonic motionA. Find force constant of spring abd period of oscillation B, find total energy of system and max acceleration of block C, find speed and acceleration when block is 3.5 cm from equilibrium D, If t=0 , the block is at x= - 2cm moving at positive x direction. Find position of block at t= 0.150 sec A 0.6 kg block is on a frictionless horizontal surface and is attached to a light spring. A horizontal force of 4.25 N is required to hold the block at rest when pulled 5cm from its equilibrium position. The block is released and undergo simple harmonic motion
A. Find force constant of spring abd period of oscillation B, find total energy of system and max acceleration of block C, find speed and acceleration when block is 3.5 cm from equilibrium D, If t=0 , the block is at x= - 2cm moving at positive x direction. Find position of block at t= 0.150 sec
A. Find force constant of spring abd period of oscillation B, find total energy of system and max acceleration of block C, find speed and acceleration when block is 3.5 cm from equilibrium D, If t=0 , the block is at x= - 2cm moving at positive x direction. Find position of block at t= 0.150 sec
Explanation / Answer
Here ,
masss , m = 0.6 Kg
at x = 5 cm = 0.05 m
force , F = 4.25 N
a) let the force constant is k
k * x = F
k * 0.05 = 4.25
k = 85 N/m
the force constant of spring is 85 N/m
period = 2pi * sqrt(m/k)
period = 2pi * sqrt(0.6/85)
period = 0.53 s
the period of oscillation is 0.53 s
b)
Total energy = 0.5 * k * x^2
Total energy = 0.5 * 85 * 0.05^2
Total energy = 0.1063 J
the Total energy is 0.1063 J
max acceleration = A * w^2
max acceleration = 0.05 * (2pi/0.53)^2
max acceleration = 7.02 m/s^2
the max acceleration is 7.02 m/s^2
c)
x = 3.5 cm = 0.035 m
let the speed is v
0.5 * 0.6 * v^2 + 0.5 * 85 * 0.035^2 = 0.5 * 85 * 0.05^2
solving for v
v = 0.425 m/s
the speed of block is 0.425 m/s
acceleration = 0.035 * 85/.6
acceleration = 4.96 m/s^2
d) at t = 0
phi = 23.6 degree
x = A * sin(w * t + phi)
x = 5 * sin((360/.53) * 0.15 - 23.6)
x = 4.9 cm
the position of the block is 4.9 cm
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