Question 18 A block weighing 13.0 N is attached to the lower end of a vertical s

Question

Question 18 A block weighing 13.0 N is attached to the lower end of a vertical spring (k = 235.0 N/m), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of 1.40 J as it passes through the point at which the spring s unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves above the point at which the spring is unstretched. (The maximum points above and below are not necessarily the same.) (c) Use the law of conservation of energy to determine the maximum distance the block moves below the point at which the spring is unstretched. (d) What s the amplitude of the oscillation? (e) What is the maximum kinetic energy of the block as it oscillates?

Explanation / Answer

here ,

a) period of oscillation , T = 2 pi * sqrt(m/k)

period of oscillation , T = 2pi * sqrt((13/9.8/235))

period of oscillation , T = 0.472 s

the period of oscillation is 0.472 s

b)
let tge maximum distance is h

Using conservation of energy

1.4 = 0.5 * 235 * x^2 + 13 * x

solving for x

x = 0.067 m = 6.7 cm

the block will rise by 6.7 cm

c)

for the distance below the position ,

let the distance is x

using consecration of energy

Using conservation of energy

1.4 = 0.5 * 235 * x^2 - 13 * x

solving for x

x = 0.177 m =17.7 cm

the block will rise by 17.7 cm

d)

amplitude of oscillation = (6.7 + 17.7)/2

amplitude of oscillation = 12.2 cm

the amplitude of oscillation is 12.2 cm

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