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A block of mass m= 2.8 kg is attached to a spring of spring constant k= 500 N/m. the block is pulled to an initial position x = 5 cm to the right of the equilibrium point as shown below, and released from rest. Find the speed of the block as it passes through equilibrium if the coefficient of kinetic friction between the block and the surface μk = 0.35.

A block of mass m 28 kg is attached to a spring of spring constant k 500 Nm the block is pulled to an initial position x 5 cm to the right of the equilibrium po class=

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Answer:

Explanation:

The energy stored in a spring

= 1/2 k x²

where k is spring constant and x is extension in the spring.

= .5 x 500 x .05²

= .625 J

Work done by friction = energy dissipated

= - μmg x d , μ is coefficient of friction , m is mass , d is displacement

= - .35 x 2.8 x 9.8 x .05

= - .48 J

energy of the mass when it reaches equilibrium position

= .625 - .48

= .145 J

If v be its velocity at that time

1/2 m v ² = .145

.5 x 2.8 x v² = .145

v² = .10357

v = .32 m /s

32 cm /s

Cricetus

The speed of block as it passes through the equilibrium will be:

"32 cm/s".

Mass and Velocity

According to the question,

Mass of block, m = 2.8 kg

Spring constant, k = 500 N/m

Initial position, x = 5 cm

Coefficient of kinetic friction, μk = 0.35

The energy stored will be:

= [tex]\frac{1}{2}[/tex] kx²

By substituting the values,

= 0.5 × 500 × (0.05)²  

= 0.625 J

Now,

Work done,

= -μmg × μ × m × d

= -0.35 × 2.8 × 9.8 × 0.05

= -0.48 J

When it reaches to the position of equilibrium, the energy of mass will be:

= 0.625 - 0.48

= 0.145 J

hence, the velocity will be:

→          [tex]\frac{1}{2}[/tex] mv² = 0.145

0.5 × 2.8 × v² = 0.145

                  v² = 0.10357

                   v = 0.32 m/s or,

                      = 32 cm/s  

Thus the above approach is correct.

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