Question

A 1.2 kg object hanging from a spring of force constant 300 N/m oscillates with a maximum speed of 30 cm/s. (a) What is its maximum displacement? When the object is at its maximum displacement, find (b) the total energy of the system, and (c) the gravitational potential energy. (Choose potential energy as zero when the object is in equilibrium).

Answer 1

Answer:

(A). The maximum displacement = 0.0189 m

(B). The Total energy of the system = 1.188 J

(C). The Gravitational potential energy = 0.22 J

Explanation:

*m = Mass of the object = 1.2 kg

*K = Spring Constant of the spring = 300 N/m

*v = Maximum speed with which the body oscillates = 30 cm/s

__Part(A)__

Let 'x' is the maximum displacement of the body

For this, we may use the Law of Conservation of Energy

$\dfrac{Kx^2}{2} = \dfrac{Mv^2}{2}$

$x^2=\dfrac{Mv^2}{K} \\\\x^2 = \dfrac{1.2\times 0.3^2}{300} \\\\x = \sqrt{0.0036} \\\\x = 0.0189 \ m$

__Part(B)__

Total energy of the system

$T.E. = \dfrac{Kx^2}{2}+\dfrac{Mv^2}{2} \\\\T.E. = 1.188 \ J$

__Part(C)__

Gravitational Potential Energy

The Gravitational potential energy of the system is

$E = Mgx \\\\E = 1.2\times 9.8\times 0.0189 = 0.222 \ J$

Answer 2

Answer:

The maximum displacement = 0.0189 m

The Total energy of the system = 1.188 J

The Gravitational potential energy = 0.22 J

Explanation:

Let m be = Mass of the object = 1.2 kg

As given K = Spring Constant of the spring = 300 N/m

Let v denote = Maximum speed with which the body oscillates = 30 cm/s

Let 'x' be the maximum displacement of the body

x²= mv²/k=1.2* 0.3 /300= 0.0189

Total energy= kinteric energy+potential energy =kx²/2 + mv²/2

=1.188J

The Gravitational potential energy of the system is

E=Mgx=1.2*9.8*0.0189