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olga nikolaevna [1]
3 years ago
5

A 2.3 kg block executes SHM while attached to a horizontal spring of spring constant 150 N/m. The maximum speed of the block as

it slides on a horizontal frictionless surface is 2.3 m/s. What are (a) the amplitude of the block's motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete 4.1 cycles of its motion?
Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer

given,

mass of block = m = 2.3 Kg

spring constant = k = 150 N/m

speed = 2.3 m/s

a) we know,

   \omega = \sqrt{\dfrac{k}{m}}

   \omega = \sqrt{\dfrac{150}{2.3}}

             ω = 8.08 rad/s

      v = A ω

      2.3 = A x 8.08

         A = 0.285 m

b) maximum acceleration

        a = A ω²

        a = 0.285 x 8.08²

        a = 18.58 m/s²

c) acceleration is minimum at mean position the minimum value of acceleration is zero.

d) time period

   T = \dfrac{2\pi}{\omega}

   T = \dfrac{2\pi}{8.08}

            T = 0.778 s

        t = 4.1 x 0.7778

        t = 3.188 s

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According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

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As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

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K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}

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