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photoshop1234 [79]
3 years ago
13

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

ilibrium point and released. Determine
a) the spring constant
b) the maximum velocity of the mass
c) the maximum acceleration of the mass
d) the total mechanical energy of the mass
e) the period and frequency of the mass and spring and
f) the equation of time-dependent vertical position of the mass
Physics
1 answer:
DochEvi [55]3 years ago
7 0

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s

Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

T=0.78sec

Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

mx'' + kx' =0

Where

'= signify order of differentiation

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