Explanation Given that a 0.2 kg mass is suspended from a vertically hanging spring, and it stretches from original length 4.5 cm to 6.0 cm . Write the extension of the spring in the vertical direction. x = 6.0 cm − 4.5 cm =1 .5 cm × 1 m 10 2 cm = 1.5 × 10 − 2 m (I) Write the expression for force experienced by the mass in the vertically downward direction. F = m g (II) Here, F is the force experienced by the mass, m is the mass, and g is the acceleration due to gravity. Write the expression for spring constant from Hook’s law. k = F x (III) Here, k is the spring constant, F is the force, and x is the extension of the spring in the vertical direction. Use equation (I) in (II). k = m g x Use equation (I) in the above equation. k = m g 1.5 × 10 − 2 m (IV) The block has a maximum speed, when the spring is at unstretched length. From conservation of energy principle, the energy associated with the spring when compressed is equal to the kinetic energy of the block 1 2 k x ′ 2 = 1 2 m v 2 Here, k is the spring constant, x ′ is the spring extension on the horizontal direction, m is the mass of the spring, and v is the maximum speed of the block. Rearrange the above, to find v . k x ′ 2 = m v 2 v 2 = k x ′ 2 m v = k x ′ 2 m (V) Write the extension of the spring in the horizontal direction

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Author: GIAMBATTISTA

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The potential energy is the energy possessed by a body by virtue of its position or the energy held by the object due to its position relative to other objects, its force like stresses, charge, and other factors affecting the particles of the body. The p…

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Chapter 6, Problem 123P

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Chapter 6, Problem 122P

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