Question

A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is

Answer 1

Answer:

a)  K_e = 0.1225 J, b)  U = 1.96 J, c) v = 0.99 m / s

Explanation:

Let's use the simple harmonium movement expression

             y = A cos (wt + Ф)

indicate that the amplitude is

             A = 0.05 m

as the system is released, the velocity at the initial point is zero

            v = dy / dt

            v = - A w sin (wt + Ф)

for t = 0 s   and v = 0 m/s

            0 = - A w sin Ф

so Ф = 0

the expression of the movement is

             y = 0.05 cos wt

The total energy of the system is

              Em = ½ k A²

let's use conservation of energy

starting point. Spring if we stretch and we set the zero of our system at this point

          Em₀ = K_e + U

          Em₀ = 0

final point. When weight and elastic force are in balance

          Em_f = K_e + U

          Em_f = ½ k y² + m g (-y)

energy is conserved

           Em₀ = Em_f

           0 = ½ k y² + m g (-y)

           k = 2mg / y

           k = 2 4.00 9.8 / 0.050

           k = 98 N / m

a) maximum elastic energy

           K_e = ½ k A²

           K_e = ½ 98 0.05²

           K_e = 0.1225 J

b) the maximum gravitational energy

            U = m g y

             U = 4.00 9.8 0.05

             U = 1.96 J

c) The maximum kinetic energy occurs when the spring is not stretched

             U = K

              mg h = ½ m v²

               v = √2gh

               v = √( 2 9.8 0.05)

               v = 0.99 m / s

d) energy at any point

               Em = K + U