0.96 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 310 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.9 s is closest to: 0.64 m/s
-0.90 m/s
0.90 m/s
-0.64 m/s
zero

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-09-02T07:33:36+02:00

Answer:

The velocity of the block is 0.64 m/s.

(A) is correct option.

Explanation:

Given that,

Mass of block = 0.96 kg

Force constant = 310 N/m

Amplitude = 0.080 m

Time = 0.9

We need to calculate the velocity of the block

We know that,

Equation of motion

[tex]x(t)=A\cos(\omega t)[/tex]

On differentiating

[tex]\dfrac{d(x(t))}{dt}=-0.080\omega \sin(\omega t)[/tex]

[tex]v(t)=-0.080\omega \sin(\omega t)[/tex]....(I)

We need to calculate the [tex]\omega[/tex]

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{310}{0.96}}[/tex]

[tex]\omega=17.96\ rad/s[/tex]

Put the value of [tex]\omega[/tex] in equation (I)

[tex]v(t)=-0.080\times17.97 \sin(17.97\times0.9)[/tex]

[tex]v(t)=0.64\ m/s[/tex]

Hence, The velocity of the block is 0.64 m/s.