A 2 g block attached to an ideal spring with spring constant 80 N/m oscillates on a horizontal friction less surface. When the spring is 4 cm shorter than its equilibrium length, the speed of block is root 17 m/s. The greatest speed of the the block is ?

Question

Answer

Given Information:

Mass of block = 2 g = 0.002 kg

Spring constant = 80 N/m

Compression of spring = 4 cm = 0.04 m

Speed of block at compression = $\sqrt{17}$ m/s

Calculation:

Find the amplitude of motion:

Using the formula for the energy of a simple harmonic oscillator, we have:

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2$

where k is the spring constant, A is the amplitude of motion, m is the mass of the block, and v is the speed of the block

Substituting the given values, we get:

$\frac{1}{2}(80)(A^2) = \frac{1}{2}(0.002)(\sqrt{17})^2$

$A = 0.05$ m

Find the maximum speed of the block:

Using the formula for the maximum speed of a simple harmonic oscillator, we have:

$v_{max} = A\sqrt{k/m}$

Substituting the given values, we get:

$v_{max} = 0.05\sqrt{\frac{80}{0.002}}$

$v_{max} = 20$ m/s

Answer:

The greatest speed of the block is 20 m/s.

Answer

Work done by spring force = ∆K.E
1/2(k)x^2=1/2m(v^2-u^2)
1/2(80) (0.04) ^2=1/2(0.002) (v^2-(√17) ^)
after solving we get v=9m/s

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