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Arlecino [84]
3 years ago
12

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

to the right of equilibrium and released from rest. Find the speed the block has as it passes through equilibrium if the coefficient of friction between block and surface is 0.350.
Physics
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position x_{i}=5.00\ cm

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx

Put the value into the formula

\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}

v^2=\dfrac{2\times12.3285}{1.00}

v^2=24.657

v=4.96\ m/s

Hence, The speed of the block is 4.96 m/s.

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The resultant vector is 5.2 cm at a direction of 12⁰ west of north.

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The resultant of the two vectors is calculated as follows;

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R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)

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Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.

Learn more about resultant vector here: brainly.com/question/28047791

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