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sveticcg [70]
3 years ago
10

A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 N/m. T

he block is pulled from its equilibrium position at x = 0.00 m to a displacement x = + 0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = -0.052 m, find the acceleration of the block.
Physics
2 answers:
Sindrei [870]3 years ago
8 0

Answer:

The block will accelerate at 92.86m/s²

Explanation:

The acceleration of a simple harmonic motion of a spring is expressed as

a= - kx/m

Where k = spring constant

x= displacement

m= mass of block

Given data

Spring constant k = 500N/m

Displacement x= - 0.052m

Mass of block m= 0.28kg

Pluging this parameters into the expression for acceleration we have

a= - 500*(-0.052)/0.28

a= 26/0.28

a= 92.86m/s²

In-s [12.5K]3 years ago
4 0

Answer:

The block will accelerate at 92.86m/s²

Explanation:

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With detailed explaniation
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  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

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\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

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\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

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