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Arte-miy333 [17]
3 years ago
9

A block on a rough, horizontal surface is attached to a horizontal spring of negligible mass. The other end of the spring is att

ached to a wall. The spring is
compressed such that the block is located at position X. When the block-spring system is released, the block travels to the right through position Y and
continues to travel to the right through position Z. Free body diagrams for the block at positions X, Y, and Z are shown in the figure. At which position does the
block have the greatest kinetic energy?
A
х
B
Y
Z
The answer cannot be determined without knowing the exact speed of the block at each position.

Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

Z

Explanation:

It has the least potential energy in the spring because the spring expanded all the way meaning it is going the fastest

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Engineers can determine properties of a structure that is modeled as a damped spring oscillator, such as a bridge, by applying a
Kazeer [188]

The value of spring constant and the oscillator's damping constant is

K= 6605.667008, b= 0.002884387

Explanation:

For Weakly damping spring oscillator

K/m = W_0^2     (at resonance)

K= mW_0^2

=0.206 * ( 2π * 28.5) ^2

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=F/W_0 * m

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8 0
3 years ago
At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl
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Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

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