Abody of mass 1kg suspended from the free end of a spring having force constant 400n/m is executing s. h.m when the total energy of system is 2 joule find the minimum acceleration

-30 m/s²

Explanation:

The total energy is the sum of the gravitational potential energy of the mass, the kinetic energy of the mass, and the elastic energy in the spring.

∑E = PE + KE + EE

∑E = mgh + ½ mv² + ½ kx²

At the bottom, h = 0 and v = 0.

∑E = ½ kx²

2 = ½ (400) x²

2 = 200 x²

0.01 = x²

x = 0.1

At the top, v = 0 and x = 0.1 − h.

∑E = mgh + ½ kx²

2 = (1) (10) h + ½ (400) (0.1 − h)²

2 = 10h + 200 (0.01 − 0.2h + h²)

2 = 10h + 2 − 40h + 200h²

0 = 200h² − 30h

0 = 10h (20h − 3)

h = 0.15

So x ranges from -0.05 m to 0.10 m.

Sum of forces on the mass:

∑F = ma

kx − mg = ma

The minimum acceleration is when x is a minimum.

(400 N/m) (-0.05 m) − (1 kg) (10 m/s²) = (1 kg) a

a = -30 m/s²

The maximum acceleration is when x is a maximum.

(400 N/m) (0.1 m) − (1 kg) (10 m/s²) = (1 kg) a

a = 30 m/s²

And of course, acceleration is 0 when kx = mg.

(400 N/m) x = (1 kg) (10 m/s²)

x = 0.025 m

see earlier