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velikii [3]
3 years ago
10

In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo

ck is held with the spring stretched 14 cm from its hanging equilibrium position. At t = 0 the block is let go. After the block evolves dynamically for 0.9 s, an ideal measurement is made of the block's speed. What is the result of this measurement (in cm/s)? Answer to one decimal place.
Physics
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

- The equation for the calculation of the period in a spring-masss system

T = \sqrt{\frac{m}{k} }     ( 1 )

- The equation for the velocity of a simple harmonic motion

x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)   ( 2 )

where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s

and by reeplacing it in ( 2 ):

v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi  sin(9\pi ) = 0

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.

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A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
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Answer:

h=12.41m

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N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

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moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

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Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

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Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

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